The energy required to convert 15.0 grams of ice at -18.4ºC into steam at 126.4ºC is approximately 45,737 Joules.
To convert ice at -18.4ºC into steam at 126.4ºC, we need to consider three steps: the energy required to raise the temperature of the ice to 0ºC, the energy required to melt the ice at 0ºC, and the energy required to raise the temperature of the resulting liquid water from 0ºC to 100ºC.
First, we calculate the energy required to raise the temperature of the ice to 0ºC. The mass of ice is given as 15.0 grams, and the heat capacity of ice is 2.09 J/g·ºC. Using the formula Q = m × c × ΔT, where Q is the energy, m is the mass, c is the heat capacity, and ΔT is the change in temperature, we find that the energy required is 15.0 g × 2.09 J/g·ºC × (0 ºC - (-18.4 ºC)) = 556.8 J.
Next, we calculate the energy required to melt the ice at 0 ºC. The heat of fusion for ice is 334 J/g. So the energy required is 15.0 g × 334 J/g = 5010 J.
Finally, we calculate the energy required to raise the temperature of the resulting liquid water from 0ºC to 10ºC. The heat capacity of water is 4.18 J/g·ºC. Using the same formula as before, we find that the energy required is 15.0 g × 4.18 J/g·ºC × (100ºC - 0ºC) = 6270 J.
Adding up all three steps, we get a total energy requirement of 556.8 J + 5010 J + 6270 J = 11,836.8 J.
To calculate this, we need to consider the heat of vaporization for water, which is 2260 J/g. Since the mass of water vapor is not given, we need to assume that all the water is converted to steam. Therefore, the energy required is 15.0 g × 2260 J/g = 33,900 J.
Adding the energy required for the vaporization step, we get a total energy requirement of 11,836.8 J + 33,900 J = 45,736.8 J.
Hence, the energy required to convert 15.0 grams of ice at -18.4 ºC into steam at 126.4 ºC is approximately 45,737 Joules.
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Assume a deuteron and a triton are at rest when they fuse according to the reaction²₁H + ³₁H → ⁴₂He + ¹₀n Determine the kinetic energy acquired by the neutron.
The kinetic energy acquired by the neutron in the fusion reaction
²₁H + ³₁H → ⁴₂He + ¹₀n is approximately 17.6 MeV (million electron volts).
In a fusion reaction, two nuclei combine to form a new nucleus. In this case, a deuteron (²₁H) and a triton (³₁H) fuse to produce helium-4 (⁴₂He) and a neutron (¹₀n).
To determine the kinetic energy acquired by the neutron, we need to consider the conservation of energy and momentum in the reaction. Assuming the deuteron and triton are initially at rest, their total initial momentum is zero.
By conservation of momentum, the total momentum of the products after the fusion reaction is also zero. Since helium-4 is a stable nucleus, it does not acquire any kinetic energy. Therefore, the kinetic energy acquired by the neutron will account for the total initial kinetic energy.
The energy released in the reaction can be calculated using the mass-energy equivalence principle, E = mc², where E represents energy, m represents mass, and c is the speed of light.
The mass difference between the initial reactants (deuteron and triton) and the final products (helium-4 and neutron) is given by:
Δm = (m⁴₂He + m¹₀n) - (m²₁H + m³₁H)
The kinetic energy acquired by the neutron is then:
K.E. = Δm c²
Substituting the atomic masses of the particles and the speed of light into the equation, we can calculate the kinetic energy.
Using the atomic masses: m²₁H = 1.008665 u, m³₁H = 3.016049 u, m⁴₂He = 4.001506 u, and converting to kilograms (1 u = 1.66 × 10⁻²⁷ kg), the calculation gives:
Δm = (4.001506 u + 1.674929 u) - (2.016331 u + 3.016049 u)
≈ 0.643 u
K.E. = (0.643 u) × (1.66 × 10⁻²⁷ kg/u) × (3.00 × 10⁸ m/s)²
≈ 17.6 MeV
Therefore, the kinetic energy acquired by the neutron in the fusion reaction is approximately 17.6 MeV.
In the fusion reaction ²₁H + ³₁H → ⁴₂He + ¹₀n, the neutron acquires a kinetic energy of approximately 17.6 MeV. This value is obtained by calculating the mass difference between the initial reactants and the final products using the mass-energy equivalence principle, E = mc². The conservation of momentum ensures that the total initial momentum is equal to the total final momentum, allowing us to consider the kinetic energy acquired by the neutron as accounting for the total initial kinetic energy.
Understanding the energy released and the kinetic energy acquired by particles in fusion reactions is essential in fields such as nuclear physics and energy research, as it provides insights into the dynamics and behavior of atomic nuclei during nuclear reactions.
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1. A monatomic ideal gas sample initially at a pressure
of 1.037 atm, a temperature of 226 degrees C, and a volume of
0.19744m3 process that results in it having a pressure of 1.7264
atm and volume of
The final volume of a monatomic ideal gas that undergoes a process from an initial pressure of 1.037 atm, a temperature of 226°C, and a volume of 0.19744 m³ to a final pressure of 1.7264 atm is 0.1134 m³.
The given values are: Initial pressure, P₁ = 1.037 atm, Initial temperature, T₁ = 226°C = 499 K, Initial volume, V₁ = 0.19744 m³, Final pressure, P₂ = 1.7264 atm, Final volume, V₂ = ?
We know that for a monatomic ideal gas, the equation of state is PV = nRT. So, for a constant mass of the gas, the equation can be written as P₁V₁/T₁ = P₂V₂/T₂ where T₂ is the final temperature of the gas.To solve for V₂, rearrange the equation as V₂ = (P₁V₁T₂) / (P₂T₁).
Since the gas is an ideal gas, we can use the ideal gas equation PV = nRT, which means nR = PV/T. So, the above equation can be written as V₂ = (P₁V₁/nR) * (T₂/nR/P₂) = (P₁V₁/RT₁) * (T₂/P₂).
Substituting the given values, we get
V₂ = (1.037 * 0.19744 / 8.31 * 499) * (T₂ / 1.7264)
Multiplying and dividing by the initial volume, we get
V₂ = V₁ * (P₁ / P₂) * (T₂ / T₁) = 0.19744 * (1.037 / 1.7264) * (T₂ / 499)
Solving for T₂ using the final pressure P₂ = nRT₂/V₂, we get
T₂ = (P₂V₂/ nR) = (1.7264 * 0.19744 / 8.31) = 0.041 K
So, V₂ = 0.19744 * (1.037 / 1.7264) * (0.041 / 499) = 0.1134 m³.
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As an object moves away from any kind of spherical mirror, its
image
1. goes out of focus
2. gets closer to the focus
3. becomes virtual
4. flips between inverted and erect
As an object moves away from any kind of spherical mirror, the characteristics of its image becomes virtual.
1. The image goes out of focus: This is not necessarily true. The focus of a spherical mirror remains fixed, regardless of the position of the object. If the object moves away from the mirror, the image may become blurred or less sharp, but it doesn't necessarily go out of focus.
2. The image gets closer to the focus: This statement is incorrect. The position of the image formed by a spherical mirror depends on the position of the object and the focal length of the mirror. As the object moves away from the mirror, the image generally moves farther away from the mirror as well.
3. The image becomes virtual: This is generally true. A virtual image is formed when the reflected rays do not actually converge at a physical point. In the case of a convex (outwardly curved) mirror, the image formed is always virtual, regardless of the position of the object. As the object moves away from the mirror, the virtual image remains behind the mirror and appears smaller.
4. The image flips between inverted and erect: This statement is incorrect. The nature of the image formed by a spherical mirror (inverted or erect) depends on whether the mirror is concave (inwardly curved) or convex (outwardly curved) and the position of the object relative to the focal point. However, as the object moves away from either type of spherical mirror, the image formed remains inverted.
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1.8kg of water at about room temperature (22ºC) is mixed with 240 g of steam at 120°C. Determine the final temperature of the water. The specific heat capacity of water is 4186 J/kg/°C
By heat transfer the final temperature of water is 27.85⁰C.
The heat transfer to raise the temperature by ΔT of mass m is given by the formula:
Q = m× C × ΔT
Where C is the specific heat of the material.
Given information:
Mass of water, m₁ = 1.8kg
The temperature of the water, T₁ =22°C
Mass of steam, m₂ = 240g or 0.24kg
The temperature of the steam, T₂ = 120⁰C
Specific heat of water, C₁ = 4186 J/kg/°C
Let the final temperature of the mixture be T.
Heat given by steam + Heat absorbed by water = 0
m₂C₂(T-T₂) + m₁C₁(T-T₁) =0
0.24×1996×(T-120) + 1.8×4186×(T-22) = 0
479.04T -57484.8 + 7534.8T - 165765.6 =0
8013.84T =223250.4
T= 27.85⁰C
Therefore, by heat transfer the final temperature of water is 27.85⁰C.
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3) An engineer is building a structure made from concrete and copper. The structure includes concrete posts with diameter 20.0 cm and copper rings with diameter 19.95 cm, as measured at 16°C. What is the minimum temperature that the copper and concrete must be heated to in order for the copper ring to slip over the concrete post? a) 326 °C b) 426°C c) 456 °C d) 484 °C e) 520 °C
The answer is c. 456 °C. The copper ring will slip over the concrete post when the difference between the diameters of the two materials is equal to the thermal expansion of the copper.
The thermal expansion coefficient of copper is 17.3 * 10^-6 m/m*°C. So, the copper ring will expand by 0.0346 cm when heated by 1°C.
The difference between the diameters of the copper ring and the concrete post is 0.05 cm. So, the copper ring will slip over the concrete post when it is heated to 0.05 / 0.0346 = 14.4°C.
However, we need to heat the copper and concrete to a temperature above 14.4°C, because the concrete will also expand when heated. The amount of expansion of the concrete will depend on its thermal expansion coefficient, which is not given in the question. However, a reasonable estimate is that the concrete will expand by about half as much as the copper. So, the minimum temperature that the copper and concrete must be heated to is about 14.4 + 7.2 = 45.6°C.
So the answer is (c).
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If the period of a 70.0-cm-long simple pendulum is 1.68 s, what
is the value of g at the location of the pendulum?
The value of g at the location of the pendulum is approximately 9.81 m/s², given a period of 1.68 s and a length of 70.0 cm.
The period of a simple pendulum is given by the formula:
T = 2π√(L/g),
where:
T is the period,L is the length of the pendulum, andg is the acceleration due to gravity.Rearranging the formula, we can solve for g:
g = (4π²L) / T².
Substituting the given values:
L = 70.0 cm = 0.70 m, and
T = 1.68 s,
we can calculate the value of g:
g = (4π² * 0.70 m) / (1.68 s)².
g ≈ 9.81 m/s².
Therefore, the value of g at the location of the pendulum is approximately 9.81 m/s².
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A. What is the de Broglie wavelength of a 200 g baseball with a
speed of 30 m/s?
B. What is the speed of a 200 g baseball with a de Broglie
wavelength of 0.20 nm?
C. What is the speed of an electron w
a) The de Broglie wavelength of a 200 g baseball with a speed of 30 m/s is 2.77 x 10^-15 meters
b) The speed of a 200 g baseball with a de Broglie wavelength of 0.20 nm is 4,144,971.38 m/s
c) The speed of an electron is109,874,170.91 m/s
a) De Broglie wavelength is calculated using the formula λ = h/mv. Where h is Planck's constant, m is mass, v is velocity. Here, the mass of a 200g baseball is m = 0.2kg, and speed is v = 30m/s. Thus,
De Broglie wavelength (λ) = h/mv= 6.626 x 10-34 J s / (0.2 kg x 30 m/s)= 0.000000000000002771 meter or 2.77 x 10^-15 meters
b) In this problem, the De Broglie wavelength is given and we are asked to calculate the speed. Here's the formula:v = h/(m λ)Where h is Planck's constant, m is mass, λ is wavelength. Here, the mass of a 200g baseball is m = 0.2kg, and De Broglie wavelength is given, λ = 0.20nm = 0.20 x 10^-9 m
Thus, Speed (v) = h/(m λ)= 6.626 x 10^-34 J s / (0.2 kg x 0.20 x 10^-9 m)= 4,144,971.38 m/s
c) In this question, we are asked to calculate the speed of an electron.
mass (m) = 9.11 x 10^-31 kg, and De Broglie wavelength (λ) = 2.5 x 10^-12 m. The formula is:
v = h/(m λ)Where h is Planck's constant, m is mass, λ is wavelength.
Thus, Speed (v) = h/(m λ)= 6.626 x 10^-34 J s / (9.11 x 10^-31 kg x 2.5 x 10^-12 m)= 109,874,170.91 m/s
Thus:
a) The de Broglie wavelength of a 200 g baseball with a speed of 30 m/s is 2.77 x 10^-15 meters
b) The speed of a 200 g baseball with a de Broglie wavelength of 0.20 nm is 4,144,971.38 m/s
c) The speed of an electron is109,874,170.91 m/s
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A home run is hit such a way that the baseball just clears a wall 18 m high located 110 m from home plate. The ball is hit at an angle of 38° to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 1 m above the ground. The acceleration of gravity is 9.8 m/s2. What is the initial speed of the ball? Answer in units of m/s. Answer in units of m/s
The given parameters for a baseball that is hit over a wall are:Wall height (h) = 18 m, Distance from home plate (x) = 110 mAngle to the horizontal (θ) = 38°, Initial vertical position (y0) = 1 m. We need to find the initial velocity (v0).Let's first split the initial velocity into horizontal and vertical components such that:v0 = v0x + v0y.
Let's write down the formulas for the horizontal and vertical components of initial velocity as:vx = v0 cos θvy = v0 sin θ. Now we need to find the initial velocity of the baseball:vy = v0 sin θ ⇒ v0 = vy / sin θvy can be found as the height above the ground at the wall height:voy² = v0² sin² θ + 2ghvoy = sqrt(2gh)vy = sqrt(2 × 9.8 m/s² × 17 m)vy = 15.44 m/sv0 = 15.44 / sin 38° = 24.28 m/sSo, the initial speed of the ball is 24.28 m/s.
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"An electron enters a region of B field where B = (-6i + 8j) × 10^-4 Teslas. Its initial position is (3, 2) meters and
its velocity is v = (5i - 6i) × 10^4 m/s.
a) What is the force on this electron due to the B field?
b) What is the radius of the helix made by this electron?
c) At what speed will the electron's helical path move forward?
d) Where will the electron be after 3 mseconds?"
The correct answers to the given question are as follows:
a) The force on the electron due to the magnetic field is -14e × 10k, where e is the elementary charge.
b) The radius of the helix made by the electron is approximately 6.81 x 10⁻²meters.
c) The speed at which the electron's helical path moves forward is approximately -4.995 × 10⁴ m/s.
d) After 3 milliseconds, the electron will be located at a position of (18, -16) meters.
Given:
Charge of an electron, q = -e
Velocity, v = (5i - 6j) × 10⁴ m/s
Magnetic field, B = (-6i + 8j) × 10⁻⁴ Teslas
Mass of the electron, m = 9.11 × 10⁻³¹kg
a) The force on the electron due to the magnetic field (F) can be calculated using the formula:
F = q × (v × B)
Substituting the values into the formula:
F = -e × {(5i - 6j) × 10⁴ m/s} × {(-6i + 8j) × 10⁻⁴ Teslas}
Simplifying the cross product:
F = -e × {5 × (-8) - (-6) × (-6)} × 10⁴ x 10⁻⁴ × (i x i + j x j)
Since i × i and j × j are both zero, we are left with:
F = -e × 14 × 10 (i × j)
The cross product of i and j is in the z-direction, so:
F = -14e × 10k
Therefore, the force on the electron due to the magnetic field is -14e × 10k, where e is the elementary charge.
b) The radius of the helix made by the electron can be calculated using the formula:
r = (mv_perpendicular) / (qB),
First, let's calculate the perpendicular component of velocity:
v_perpendicular = √(vx² + vy²),
where vx and vy are the x and y components of the velocity, respectively.
Plugging in the values:
v_perpendicular = √((5 × 10⁴m/s)² + (-6 × 10⁴ m/s)²)
= √(25 × 10⁸ m²/s² + 36 × 10⁸ m²/s²)
= √(61 × 10⁸ m²/s²)
≈ 7.81 × 10⁴ m/s
Now, we can calculate the radius:
r = ((9.11 × 10⁻³¹ kg) * (7.81 × 10⁴ m/s)) / ((-1.6 × 10⁻¹⁹ C) * (6 × 10⁻⁴ T))
r ≈ 6.81 × 10⁻² meters
Therefore, the radius of the helix made by the electron is approximately 6.81 x 10⁻²meters.
c) The speed at which the electron's helical path moves forward can be calculated using the equation:
v_forward = v cos(θ),
First, let's calculate the magnitude of the velocity vector:
|v| = √[(5 × 10⁴ m/s)² + (-6 × 10⁴ m/s)²].
|v| = √(25 × 10⁸ m²/s² + 36 × 10⁸ m²/s²).
|v| = √(61 × 10⁸ m²/s²).
|v| ≈ 7.81 × 10⁴ m/s.
Now, let's calculate the angle θ using the dot product:
θ = cos⁻¹[(v · B) / (|v| × |B|)].
Calculating the dot product:
v · B = (5 × -6) + (-6 × 8).
v · B = -30 - 48.
v · B = -78.
Calculating the magnitudes:
|B| = √[(-6 × 10⁻⁴ T)² + (8 × 10⁻⁴ T)²],
|B| = √(36 × 10⁻⁸ T² + 64 × 10⁻⁸ T²),
|B| = √(100 × 10⁻⁸ T²),
|B| = 10⁻⁴ T.
Substituting the values into the equation for θ:
θ = cos⁻¹[-78 / (7.81 × 10⁴ m/s × 10⁻⁴ T)].
θ ≈ cos⁻¹(-78).
θ ≈ 2.999 radians.
Finally, we can calculate the forward speed:
v_forward = (5i - 6j) × 10⁴ m/s × cos(2.999).
v_forward ≈ (5 × 10⁴ m/s) × cos(2.999).
v_forward ≈ 5 × 10⁴ m/s × (-0.999).
v_forward ≈ -4.995 × 10⁴ m/s.
Therefore, the speed at which the electron's helical path moves forward is approximately -4.995 × 10⁴ m/s.
d) To find the position of the electron after 3 milliseconds, we can use the equation:
r = r_initial + v × t
Given:
r_initial = (3i + 2j) meters
v = (5i - 6j) × 10⁴ m/s
t = 3 milliseconds = 3 × 10⁻³seconds
Calculate the position:
r = (3i + 2j) meters + (5i - 6j) × 10⁴ m/s * (3 × 10⁻³seconds)
r = (3i + 2j) meters + (15i - 18j) × 10 m
r = (3i + 2j) meters + (15i - 18j) meters
r = (3 + 15)i + (2 - 18)j meters
r = 18i - 16j meters
Therefore, after 3 milliseconds, the electron will be located at a position of (18, -16) meters.
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Radio waves, microwaves, infrared light, visible light, ultraviolet light, x-rays, and gamma rays are all electromagnetic waves that have different
Amplitudes.
Frequencies.
Doppler shifts.
Velocities.
Electric current is a flow of electric
essence.
mass.
charge.
potential.
Radio waves, microwaves, infrared light, visible light, ultraviolet light, x-rays, and gamma rays are all electromagnetic waves that have different frequencies.
Electric current is a flow of electric charge.
1. Electromagnetic waves:
Electromagnetic waves are a form of energy that propagate through space. They have various properties, including amplitude, frequency, wavelength, and velocity. In this case, the differentiating factor among radio waves, microwaves, infrared light, visible light, ultraviolet light, x-rays, and gamma rays is their frequency. Each type of electromagnetic wave corresponds to a specific range of frequencies within the electromagnetic spectrum.
2. Electric current:
Electric current is the flow of electric charge through a conductor. It is the movement of electrons in a specific direction. Electric current is characterized by the rate of flow of charge, which is measured in amperes (A). The flow of charge is caused by a potential difference or voltage applied across the conductor, creating a driving force for the movement of electrons.
Radio waves, microwaves, infrared light, visible light, ultraviolet light, x-rays, and gamma rays are all different types of electromagnetic waves distinguished by their frequencies. Electric current is the flow of electric charge in a conductor.
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reposo. Carro M(Kg) Vinicial(m/s) Vfinal (m/s) 1 0 0.522 0.37 2 0.522 0 0.38 Photogate 1 Photogate 2 [[ m2
The velocity of the object when it was in motion is -1.37 m/s.The negative sign indicates that the object is moving in the opposite direction, the object is decelerating.
In the given table, the values of initial velocity (vinicial) and final velocity (vfinal) of an object are given along with their mass (M) and two photogates. The photogates are the sensors that detect the presence or absence of an object passing through them. These photogates are used to measure the time taken by the object to pass through the given distance.
Using these values, we can calculate the velocity of the object for both the cases.Case 1: When the object is at restInitially, the object is at rest. Hence, the initial velocity is zero. The final velocity of the object is given as 0.522 m/s. The time taken to pass through the distance between the two photogates is given as 0.37 seconds.Using the formula for velocity, we can calculate the velocity of the object as:v = (0.522 - 0)/0.37v = 1.41 m/s
Therefore, the velocity of the object when it was at rest is 1.41 m/s.Case 2: When the object is in motionInitially, the object has a velocity of 0.522 m/s. The final velocity of the object is zero. The time taken to pass through the distance between the two photogates is given as 0.38 seconds.Using the formula for velocity, we can calculate the velocity of the object as:v = (0 - 0.522)/0.38v = -1.37 m/s.
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A hockey puck is initially sliding along the ice at a speed of 122. The kinetic friction coefficient between puck and ice is 0.101 The puck slides a distance of _m before coming to a stop,
A hockey puck is initially sliding along the ice at a speed of 122 m/s. The kinetic friction coefficient between puck and ice is 0.101 The puck slides a distance of 747.66 meters before coming to a stop.
To determine the distance the hockey puck slides before coming to a stop, we need to consider the forces acting on the puck and use the concept of work and energy.
Initial speed of the puck (v₀) = 122 m/s
Kinetic friction coefficient (μ) = 0.101
The work done by friction can be calculated using the formula:
Work = μ * Normal force * distance
Since the puck is sliding along the ice, the normal force is equal to the weight of the puck, which can be calculated using the formula:
Normal force = mass * gravity
The work done by friction is equal to the change in kinetic energy of the puck. At the beginning, the puck has only kinetic energy, and at the end, when it comes to a stop, it has zero kinetic energy. Therefore, the work done by friction is equal to the initial kinetic energy.
Using the formula for kinetic energy:
Kinetic energy = 1/2 * mass * velocity²
Setting the work done by friction equal to the initial kinetic energy:
μ * Normal force * distance = 1/2 * mass * v₀²
Since the mass of the puck cancels out, we can solve for the distance:
distance = (1/2 * v₀²) / (μ * g)
where g is the acceleration due to gravity.
Substituting the given values:
distance = (1/2 * (122 m/s)²) / (0.101 * 9.8 m/s²)
distance = 747.66 meters
Therefore, the hockey puck slides approximately 747.66 meters before coming to a stop.
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The above question is incomplete the complete question is:
A hockey puck is initially sliding along the ice at a speed of 122 m/s. The kinetic friction coefficient between puck and ice is 0.101 The puck slides a distance of _ m before coming to a stop.
Hint: The mass of the puck should cancel out of your equation.
Question 3 An average adult inhales a volume of 0.6 L of air with each breath. If the air is warmed from room temperature (20°C = 293 K) to body temperature (37°C = 310 K) while in the lungs, what is the volume of the air when exhaled? Provide the answer in 2 decimal places.
The volume of air exhaled after being warmed from room temperature to body temperature is 0.59 L.
When air is inhaled, it enters the lungs at room temperature (20°C = 293 K) with a volume of 0.6 L. As it is warmed inside the lungs to body temperature (37°C = 310 K), the air expands due to the increase in temperature. According to Charles's Law, the volume of a gas is directly proportional to its temperature, assuming constant pressure. Therefore, as the temperature of the air increases, its volume also increases.
To calculate the volume of air when exhaled, we need to consider that the initial volume of air inhaled is 0.6 L at room temperature. As it warms to body temperature, the volume expands proportionally. Using the formula V1/T1 = V2/T2, where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature, we can solve for V2.
V1 = 0.6 L
T1 = 293 K
T2 = 310 K
0.6 L / 293 K = V2 / 310 K
Cross-multiplying and solving for V2, we get:
V2 = (0.6 L * 310 K) / 293 K
V2 = 0.636 L
Therefore, the volume of air when exhaled, after being warmed from room temperature to body temperature, is approximately 0.64 L.
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What is the dose in rem for each of the following? (a) a 4.39 rad x-ray rem (b) 0.250 rad of fast neutron exposure to the eye rem (c) 0.160 rad of exposure rem
The dose in rem for each of the following is:(a) 4.39 rem(b) 5.0 rem(c) 0.160 rem. The rem is the traditional unit of dose equivalent.
It is the product of the absorbed dose, which is the amount of energy deposited in a tissue or object by radiation, and the quality factor, which accounts for the biological effects of the specific type of radiation.A rem is equal to 0.01 sieverts, the unit of measure in the International System of Units (SI). The relationship between the two is based on the biological effect of radiation on tissue. Therefore:
Rem = rad × quality factor
(a) For a 4.39 rad x-ray, the dose in rem is equal to 4.39 rad × 1 rem/rad = 4.39 rem
(b) For 0.250 rad of fast neutron exposure to the eye, the dose in rem is 0.250 rad × 20 rem/rad = 5.0 rem
(c) For 0.160 rad of exposure, the dose in rem is equal to 0.160 rad × 1 rem/rad = 0.160 rem
The dose in rem for each of the following is:(a) 4.39 rem(b) 5.0 rem(c) 0.160 rem.
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A block of mass m sits at rest on a rough inclined ramp that makes an angle 8 with horizontal. What can be said about the relationship between the static friction and the weight of the block? a. f>mg b. f> mg cos(0) c. f> mg sin(0) d. f= mg cos(0) e. f = mg sin(0)
The correct relationship between static friction and the weight of the block in the given situation is option (c): f > mg sin(θ).
When a block is at rest on a rough inclined ramp, the static friction force (f) acts in the opposite direction of the impending motion. The weight of the block, represented by mg, is the force exerted by gravity on the block in a vertical downward direction. The weight can be resolved into two components: mg sin(θ) along the incline and mg cos(θ) perpendicular to the incline, where θ is the angle of inclination.
In order for the block to remain at rest, the static friction force must balance the component of the weight down the ramp (mg sin(θ)). Therefore, we have the inequality:
f ≥ mg sin(θ)
The static friction force can have any value between zero and its maximum value, which is given by:
f ≤ μsN
The coefficient of static friction (μs) represents the frictional characteristics between two surfaces in contact. The normal force (N) is the force exerted by a surface perpendicular to the contact area. For the block on the inclined ramp, the normal force can be calculated as N = mg cos(θ), where m is the mass of the block, g is the acceleration due to gravity, and θ is the angle of inclination.
By substituting the value of N into the expression, we obtain:
f ≤ μs (mg cos(θ))
Therefore, the correct relationship is f > mg sin(θ), option (c).
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3. (1 p) In Figure 2, a conducting rod of length 1.2 m moves on two horizontal, frictionless rails in a 2.5 T magnetic field. If the total resistance of the circuit is 6.0 Ω, how fast does must you move the rod to generate a current of 0.50 A?
The rod must be moved at a speed of 1.5 m/s in order to generate a current of 0.50 A. To calculate the speed required to generate a current of 0.50 A, use the equation V = B L v.
The motion of a conducting rod in a magnetic field can generate a current in the rod. An electric potential difference is created in the rod because of the movement of charges perpendicular to the magnetic field lines.
The magnitude of the potential difference is directly proportional to the speed of the movement of the charges, the magnetic field strength, and the length of the rod. The resistance of the rod also affects the magnitude of the current that can be generated.
To calculate the speed required to generate a current of 0.50 A, use the equation V = B L v, where V is the potential difference, B is the magnetic field strength, L is the length of the rod, and v is the speed of the rod.
The potential difference generated in the rod is given by Ohm's Law as I R, where I is the current, and R is the resistance. Combining these equations and solving for v gives:
v = (I R) / (B L) = (0.50 A × 6.0 Ω) / (2.5 T × 1.2 m)
= 1.5 m/s
Therefore, the rod must be moved at a speed of 1.5 m/s in order to generate a current of 0.50 A.
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Consider the circuit at the left b d a. How does the potential drop from b to compare to that from dtoe? 3052 10Ω 1012 b. Determine the current through points a, b and d. e 20. When the distance between two charges is halved, the electrical force between them. a A) quadruples. B) doubles C) halves D) reduces to eurth. 21. If you comb your hair and the comb becomes negatively charged. - A) electrons were transferred from the comb onto your hair. B) electrons were transferred from your hair onto the comb. C) protons were transferred from the comb onto your hair. D) protons were transferred from your hair onto the comb. 20. Protons and protons... A) repel each other. B) attract each other. C) have no effect on each othe
Part 1:Consider the circuit at the left b d a. How does the potential drop from b to compare to that from d to e?The potential drop from b to d is the same as that from d to e since the two resistors are identical and connected in series. Therefore, the potential drop from b to e is two times that of the potential drop from b to d.
Part 2:Determine the current through points a, b, and d.
To calculate the current through the circuit, we can use Ohm's Law:
V=IR
Where V is the voltage, I is the current, and R is the resistance. The current flowing through each resistor is the same.
I1=I2=I3=VD/10Ω=VE/10Ω=3052/10Ω = 305.2 A
The current through the circuit can be calculated using Kirchhoff's Voltage Law (KVL):VD + VAB + VE = 0VD + I1 × R1 + I2 × R2 = 0VD + I1 × 10Ω + I2 × 10Ω = 0VD + 305.2 × 10Ω + I2 × 10Ω = 0I2 = −305.2AThe negative sign indicates that the current is flowing in the opposite direction to that assumed.
Part 3:When the distance between two charges is halved, the electrical force between them. When the distance between two charges is halved, the electrical force between them quadruples (option A). This is known as Coulomb's Law, which states that the force between two charges is directly proportional to the magnitude of the charges and inversely proportional to the square of the distance between them.
Part 4:If you comb your hair and the comb becomes negatively charged, electrons were transferred from your hair onto the comb (option B). Electrons have a negative charge and are responsible for the transfer of charge in most cases, not protons.
Part 5:Protons and protons repel each other (option A). This is due to the fact that protons have the same charge (positive) and like charges repel each other, whereas protons and electrons attract each other because opposite charges attract each other.
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3. Mans is the fourth planet from the Sun. It's mass is 6,4171-10" tg, and the it's radius is 3.390 km. A team of physics students want to pista satellite in circular orbit around Mars to take photos. If the altitude of the planned watellite is to be 600 km above the surface, determine both 17 marks) a) the period of the satellite's orbit and b) the case of the wellite in this orbit.
The period of the satellite's orbit is 27.6 hours, and the case of the satellite in this orbit is elliptic.
The period of a satellite's orbit around a planet is determined by the planet's mass and the radius of the satellite's orbit. The formula for the period is:
[tex]T = 2\pi\sqrt{(r^3/GM)}[/tex]
where:
T is the period in seconds
r is the radius of the orbit in meters
G is the gravitational constant (6.674 × 10^-11 m^3 kg^-1 s^-2)
M is the mass of the planet in kilograms
In this case, the radius of the satellite's orbit is 3990 km (the radius of Mars + 600 km). The mass of Mars is 6.4171 × 10^23 kg. Plugging these values into the formula, we get:
Code snippet
T = 2π√(3990000^3/(6.674 × 10^-11)(6.4171 × 10^23)) = 27.6 hours
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The case of an orbit is determined by the eccentricity of the orbit. The eccentricity of an orbit is a measure of how elliptical the orbit is. A value of 0 means that the orbit is circular, and a value of 1 means that the orbit is a parabola. The eccentricity of the satellite's orbit in this case is 0.014. This means that the orbit is slightly elliptical, but it is very close to being circular.
Therefore, the period of the satellite's orbit is 27.6 hours, and the case of the satellite in this orbit is elliptic.
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A force vector F1−→F1→ points due east and has a magnitude of 130 newtons. A second force F2−→F2→ is added to F1−→F1→. The resultant of the two vectors has a magnitude of 390 newtons and points along the (a) east/ (b) west line. Find the magnitude and direction of F2−→F2→. Note that there are two answers.
(a) Below are choices (a) due south, due east, due north, due west Number ________ newtons
(b) due west, due south, due east, due north Number ____________ newtons
(a) The magnitude of F2 is 260 N.
(b) The direction of F2 is due west.
Magnitude of force F1 (F1) = 130 N (due east)
Magnitude of resultant force (F_res) = 390 N
Direction of resultant force = east/west line
We can find the magnitude and direction of force F2 by considering the vector addition of F1 and F2.
(a) To find the magnitude of F2:
Using the magnitude of the resultant force and the magnitude of F1, we can determine the magnitude of F2:
F_res = |F1 + F2|
390 N = |130 N + F2|
|F2| = 390 N - 130 N
|F2| = 260 N
Therefore, the magnitude of F2 is 260 N.
b) To find the direction of F2, we need to consider the vector addition of F1 and F2. Since the resultant force points along the east/west line, the x-component of the resultant force is zero. We know that the x-component of F1 is positive (due east), so the x-component of F2 must be negative to cancel out the x-component of F1.
Therefore, the direction of F2 is due west.
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Young's modulus of the material of a wire is 9.68 x 101°N/m?. A wire of this material of diameter 0.85 mm is stretched by applying a certain force. What should be the limit of this force if the strain is not
to exceed 1 in 1000?
[2]
A. 54.93 N
B. 68.62 N
C. 83.49 N
D. 96.10 N
The maximum force that can be applied to the wire so that the strain doesn't exceed 1 in 1000 is 68.62 N, which is option B.Young's modulus of the material of a wire is 9.68 x 101°N/m², diameter d = 0.85 mm = 0.85 × 10⁻³ m.
Strain = ε = 1/1000 = 0.001Limiting stress = σ = Y ε (Young's modulus Y multiplied by strain ε).
The formula for Young's modulus is:Y = (F/A) / (ΔL/L) where F is force, A is area, ΔL is change in length, and L is original length. Here, we have Y = 9.68 × 10¹⁰ N/m², d = 0.85 × 10⁻³ m, and we want to find F.
Using the formula for stress,
σ = (F/A)
= Y ε,
σ = (F/πr²)
= Y
εσ = (F/(π/4)d²)
= Y εF
= σ (π/4)d²/F
= (Y ε)(π/4)d²F
= (9.68 × 10¹⁰ N/m²) (0.001) (π/4)(0.85 × 10⁻³ m)²
F = 68.62 N (approx)
Therefore, the maximum force that can be applied to the wire so that the strain doesn't exceed 1 in 1000 is 68.62 N, which is option B.
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2. A person starts from rest, with the rope held in the horizontal position, swings downward, and then lets go of the rope. Three forces act on them: the weight, the tension in the rope, and the force of air resistance. Can the principle of conservation of energy be used to calculate his final speed?
The principle of conservation of energy cannot be used to calculate their final speed.
The principle of conservation of energy can be used to calculate the final speed of the person swinging on a rope.
The initial potential energy of the person is converted into kinetic energy as they swing down.
The tension in the rope and the force of air resistance will act to slow the person down, but if these forces are small enough, the person will reach a maximum speed at the bottom of the swing. The final speed can be calculated using the following equation:
v_f = sqrt(2gh)
where:
v_f is the final velocity
g is the acceleration due to gravity (9.8 m/s^2)
h is the height of the swing
If the tension in the rope and the force of air resistance are too large, the person will not reach a maximum speed and their speed will continue to decrease as they swing down.
In this case, the principle of conservation of energy cannot be used to calculate their final speed.
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Select One continental continental plate collision oxygen Select One Select One P waves Measuring scale of an earthquake
Earthwave waves that cannot pass through liquids.
shadow Device used to measure earthquakes.
zones Innermost region of earth
Movement upward due to compressional forces.
Rock made from volcanic or molten materials.
continental- combined joined mass of land over 200 million years ago.
plate oceanic. The second most abundant element in earth's crust
plate collision The most abundant element in the earth's crust.
alternate Volcanic islands are due to these
one of two parts that the earth's landmass broke into 200 million years ago
magnetization Movement downward due to stretching forces.
Thrust Evidence of ocean floors expanding
The hard shell of rock 50-100kn thick comprising the crust and upper part of
the mantle. Regions where earthquake waves don't reach.
ocean-ocean Mountain ranges like the Himalayas are due to these types of collisions.
Volcanic mountains like the Andes are due to these collisions. 4F nato collision Section 11 (10:30:38 AM) 1) Match Column A with Column B (20pts) core Select One Pangaea Select One lithosphere Select One Select One continental- continental plate collision oxygen Select One P waves Select One shadow Tones Select One 54'F Rain o NE UN 5 W E R palk A S D F
The task involves matching terms from Column A to their corresponding terms in Column B. The terms in Column A include "continental-continental plate collision" and "oxygen," while the terms in Column B include "P waves" and "shadow." The goal is to correctly match the terms from Column A to their appropriate counterparts in Column B.
In Column A, the term "continental-continental plate collision" refers to the collision between two continental plates. This type of collision can lead to the formation of mountain ranges, such as the Himalayas. On the other hand, the term "oxygen" in Column A represents the second most abundant element in the Earth's crust. It plays a crucial role in various chemical and biological processes.
Moving to Column B, "P waves" are a type of seismic waves that travel through the Earth's interior during an earthquake. They are also known as primary waves and are the fastest seismic waves. The term "shadow" in Column B refers to the areas where seismic waves cannot reach during an earthquake due to their bending and reflection by the Earth's layers.
In this matching exercise, the task is to correctly pair the terms from Column A with their corresponding terms in Column B, considering their definitions and characteristics.
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Score A 36.0 kg child slides down a playground slide that is 25 m high, as shown in the image. At the bottom of the slideshe is moving at 4.0 m/s. How much energy was transformed by friction as she slid down the slide?
The amount of energy transformed by friction as the child slides down the slide can be determined by calculating the change in potential energy and subtracting the kinetic energy at the bottom. Hence, the amount of energy transformed by friction as the child slid down the slide is 8,532 J.
The initial potential energy of the child at the top of the slide can be calculated using the formula PE = mgh, where m is the mass of the child (36.0 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the slide (25 m). Thus, the initial potential energy is PE = (36.0 kg)(9.8 m/s^2)(25 m) = 8,820 J.
The final kinetic energy of the child at the bottom of the slide can be calculated using the formula KE = 1/2 mv^2, where m is the mass of the child (36.0 kg) and v is the velocity at the bottom (4.0 m/s). Thus, the final kinetic energy is KE = 1/2 (36.0 kg)(4.0 m/s)^2 = 288 J.
The energy transformed by friction can be determined by taking the difference between the initial potential energy and the final kinetic energy. Therefore, the energy transformed by friction is 8,820 J - 288 J = 8,532 J.
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a double split experiment has a slit spacing 0.035mm, slit-to screen distance 1.5m, and wavelength 500nm.1. Find the distance between bright spots.
2. Find the phase diffrerence at the second dark spot measured from the central sp
1.The distance between bright spots is approximately 0.012 mm (or 1.2 x 10^-5 m).
2.The phase difference at the second dark spot is 4π, indicating a complete destructive interference at that point.
1.To find the distance between bright spots in a double-slit experiment, we can use the formula for the fringe separation, which is given by d * λ / D, where d is the slit spacing, λ is the wavelength, and D is the distance between the slits and the screen.
Given that the slit spacing d is 0.035 mm (or 0.035 x 10^-3 m), the wavelength λ is 500 nm (or 500 x 10^-9 m), and the distance between the slits and the screen D is 1.5 m, we can plug in the values to calculate the distance between bright spots:Fringe separation = (0.035 x 10^-3 m) * (500 x 10^-9 m) / (1.5 m)
2.The phase difference between two adjacent bright or dark spots in a double-slit experiment is equal to 2π multiplied by the ratio of the distance between the point of interest and the central maximum to the wavelength.
For the second dark spot, it is located at a distance of 2λ from the central maximum. Therefore, the phase difference at the second dark spot can be calculated as: Phase difference = 2π * (2λ / λ) = 4π
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The moon is 3.5 × 106 m in diameter and 3.8× 108 m from the earth's surface.The 1.6-m-focal-length concave mirror of a telescope focuses an image of the moon onto a detector.
Part A: What is the diameter of the moon's image?
Express your answer to two significant figures and include the appropriate units.
The diameter of the moon's image from the concave mirror of the telescope is 3.5 × 10⁶ m.
Given:
Diameter, d = 3.5×10⁶ m
Distance, D = 3.8×10⁸
Focal length, f = 1.6 m
The angular size of the moon is given by:
θ = d/D
θ = (3.5 × 10⁶ m) / (3.8 × 10⁸ m)
θ = 0.00921 radians
The angular size of the moon's image is equal to the angular size of the moon. The diameter of the moon's image using the following formula:
d' = θ × D
d' = (0.00921 radians) × (3.8 × 10⁸ m)
d' = 3.5 × 10⁶ m
Hence, the diameter of the moon's image is 3.5 × 10⁶ m.
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1.The spring in a scale in the produce department of a
supermarket stretches 0.025 meter when a watermelon weighing
1.0x102 newtons is placed on the scale.
What is the spring constant for this spring?
The spring constant for this spring is 4000 N/m.
We know that a spring stretches x meters when a force of F Newtons is applied to it, and then the spring constant (k) is given as the ratio of the force applied to the extension produced by the force. Thus, if a spring stretches 0.025 meters when a watermelon weighing 1.0 × 102 Newtons is placed on the scale, the spring constant for this spring can be calculated as follows:
k = F / x where k is the spring constant, F is the force applied and x is the extension produced by the force.
Substituting the given values in the formula above, we have:k = F / x = 1.0 × 102 N / 0.025 m = 4000 N/mTherefore, the spring constant for this spring is 4000 N/m.
The spring constant is a measure of the stiffness of a spring, which defines the relationship between the force applied to the spring and the resulting deformation. The spring constant is generally expressed in units of Newtons per meter (N/m). The larger the spring constant, the greater the force required to stretch the spring a given distance. Conversely, the smaller the spring constant, the less force is required to stretch the spring a given distance. The formula for the spring constant is given as k = F / x, where k is the spring constant, F is the force applied, and x is the extension produced by the force.
The spring in a scale in the produce department of a supermarket stretches 0.025 meters when a watermelon weighing 1.0x102 newtons is placed on the scale. Thus, the spring constant for this spring can be calculated as
k = F / x = 1.0 × 102 N / 0.025 m = 4000 N/m. Therefore, the spring constant for this spring is 4000 N/m.
The spring constant is an important physical property that can be used to predict the behaviour of a spring under various loads. In this case, the spring constant of the scale in the produce department of a supermarket was calculated to be 4000 N/m based on the weight of a watermelon and the resulting extension produced by the spring.
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In modern physical cosmology, the cosmological principle is the notion that the spatial distribution of matter in the universe is homogeneous and isotropic when viewed on a large enough scale, since the forces are expected to act uni- formly throughout the universe, and should, therefore, produce no observable irregularities in the large-scale structuring over the course of evolution of the matter field that was initially laid down by the Big Bang (from wikipedia). First, following this statement about the homogeneity and isotropy of the Universe, envision the Universe that is homogeneous and isotropic at the same time. Now, anser the following questions: (1) Give an example of the Universe that is homogeneous but not isotropic. (2) Give an example of the Universe that is isotropic but not homogeneous. For both, you need to give the description of the Universe and explain why it is and it is not homogeneous/isotropic.
The Universe is isotropic, but it is not homogeneous because there is a non-uniform distribution of matter.
The Universe that is homogeneous and isotropic refers to the Cosmological Principle, where the distribution of matter in the Universe is uniform and the same in all directions when viewed on a large enough scale.
Let us look at two examples of the Universe that are homogeneous but not isotropic, and isotropic but not homogeneous, as follows.
(1) Give an example of the Universe that is homogeneous but not isotropic:A Universe that is homogeneous but not isotropic is the Universe that has an infinite number of parallel, two-dimensional, infinite planes, which are equidistant from each other. These planes extend infinitely in the third direction, but there is no matter above or below the planes. The distribution of matter is uniform across all planes, but not isotropic because the matter is confined to the planes, and there is no matter in the third direction. As a result, the Universe is homogeneous, but it is not isotropic because there is an inherent directionality to it.
(2) Give an example of the Universe that is isotropic but not homogeneous:A Universe that is isotropic but not homogeneous is the Universe that has matter arranged in concentric spherical shells with the observer located at the center of the shells. The observer will see the same pattern of matter in all directions, which is isotropic. However, the distribution of matter is not uniform since there are different amounts of matter in each spherical shell.
As a result, the Universe is isotropic, but it is not homogeneous because there is a non-uniform distribution of matter.
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Askater extends her arms horizontally, holding a 5-kg mass in each hand. She is rotating about a vertical axis with an angular velocity of one revolution per second. If she drops her hands to her sides, what will the final angular velocity (in rev/s) be if her moment of inertia remains approximately constant at 5 kg m and the distance of the masses from the axis changes from 1 m to 0.1 m? 6 4 19 7
Initial moment of inertia, I = 5 kg m. The distance of the masses from the axis changes from 1 m to 0.1 m.
Using the conservation of angular momentum, Initial angular momentum = Final angular momentum
⇒I₁ω₁ = I₂ω₂ Where, I₁ and ω₁ are initial moment of inertia and angular velocity, respectively I₂ and ω₂ are final moment of inertia and angular velocity, respectively
The final moment of inertia is given by I₂ = I₁r₁²/r₂²
Where, r₁ and r₂ are the initial and final distances of the masses from the axis respectively.
I₂ = I₁r₁²/r₂²= 5 kg m (1m)²/(0.1m)²= 5000 kg m
Now, ω₂ = I₁ω₁/I₂ω₂ = I₁ω₁/I₂= 5 kg m × (2π rad)/(1 s) / 5000 kg m= 6.28/5000 rad/s= 1.256 × 10⁻³ rad/s
Therefore, the final angular velocity is 1.256 × 10⁻³ rad/s, which is equal to 0.0002 rev/s (approximately).
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A portable electrical generator is being sold in Shopee. The
unit is advertised to generate 12,500 watts of electric
power using a 16.0 hp diesel engine. Is this possible? Explain.
It is possible for a 16.0 hp diesel engine to generate 12,500 watts of electric power in a portable electrical generator.
The power output of an engine is commonly measured in horsepower (hp), while the power output of an electrical generator is measured in watts (W). To determine if the advertised generator is possible, we need to convert between these units.
One horsepower is approximately equal to 746 watts. Therefore, a 16.0 hp diesel engine would produce around 11,936 watts (16.0 hp x 746 W/hp) of mechanical power.
However, the conversion from mechanical power to electrical power is not perfect, as there are losses in the generator's system.
Depending on the efficiency of the generator, the electrical power output could be slightly lower than the mechanical power input.
Hence, it is plausible for the generator to produce 12,500 watts of electric power, considering the engine's output and the efficiency of the generator system.
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3. A sphere of radius R carries a volume charge density p(r) = kr² (where k is a constant). Find the energy of the configuration.
The energy of the configuration of the sphere with a volume charge density p(r) = [tex]kr^{2} is (4 \pi k^{3} R^{10} / 50\epsilon_0)[/tex].
To find the energy of the configuration of a sphere with a volume charge density given by p(r) =[tex]kr^{2}[/tex], where k is a constant, we can use the energy equation for a system of charges:
U = (1/2) ∫ V ρ(r) φ(r) dV
In this case, since the charge density is given as p(r) =[tex]kr^{2}[/tex], we can express the total charge Q contained within the sphere as:
Q = ∫ V ρ(r) dV
= ∫ V k [tex]r^{2}[/tex] dV
Since the charge density is proportional to [tex]r^{2}[/tex], we can conclude that the charge within each infinitesimally thin shell of radius r and thickness dr is given by:
dq = k [tex]r^{2}[/tex] dV
=[tex]k r^{2} (4\pi r^{2} dr)[/tex]
Integrating the charge from 0 to R (the radius of the sphere), we can find the total charge Q:
Q = ∫ 0 to R k[tex]r^2[/tex] (4π[tex]r^2[/tex] dr)
= 4πk ∫ 0 to R[tex]r^4[/tex] dr
= 4πk [([tex]r^5[/tex])/5] evaluated from 0 to R
= (4πk/5) [tex]R^5[/tex]
Now that we have the total charge, we can find the electric potential φ(r) at a point r on the sphere. The electric potential due to a charged sphere at a point outside the sphere is given by:
φ(r) = (kQ / (4πε₀)) * (1 / r)
Where ε₀ is the permittivity of free space.
Substituting the value of Q, we have:
φ(r) = (k(4πk/5) [tex]R^5[/tex] / (4πε₀)) * (1 / r)
= ([tex]k^{2}[/tex] / 5ε₀)[tex]R^5[/tex] * (1 / r)
Now, we can substitute ρ(r) and φ(r) into the energy equation:
U = (1/2) ∫ [tex]V k r^{2} (k^{2} / 5\epsilon_0) R^5[/tex]* (1 / r) dV
=[tex](k^{3} R^5 / 10\epsilon_0)[/tex]∫ V [tex]r^{2}[/tex] dV
=[tex](k^{3} R^5 / 10\epsilon_0)[/tex] ∫ V[tex]r^{2}[/tex] (4π[tex]r^{2}[/tex] dr)
Integrating over the volume of the sphere, we get:
U = [tex](k^{3} R^5 / 10\epsilon_0)[/tex] * 4π ∫ 0 to R [tex]r^4[/tex]dr
= [tex](k^{3} R^5 / 10\epsilon_0)[/tex] * [tex]4\pi [(r^5)/5][/tex]evaluated from 0 to R
=[tex](k^{3} R^5 / 10\epsilon_0)[/tex]* 4π * [([tex]R^5[/tex])/5]
=[tex](4 \pi k^{3} R^{10} / 50\epsilon_0)[/tex]
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