What is the net change in energy of a system over a period of 1.5 hours if the system has a power output of 140W? O A. 70.0 kJ O B. 756.0 kJ C. 93.3 kJ O D. 1.6 kJ

2048.77 inches space needed to be allowed for expansion

To calculate the expansion space required for a steel walkway that spans a 170 ft 8.77 inch gap.

we need to consider the walkway's coefficient of thermal expansion and the temperature range it's designed for. Using the given coefficient of and the temperature range of -32.4 C to 39.4 C, we can calculate the expansion space required in inches, which turns out to be 2.39 inches.

The expansion space required for the steel walkway can be calculated using the following formula:

ΔL = L * α * ΔT

Where ΔL is the change in length of the walkway, L is the original length (in this case, the length of the gap the walkway spans), α is the coefficient of thermal expansion, and ΔT is the temperature difference.

[tex]ΔL = 170 ft 8.77 in * (18.4 \times 10^-6 mm/mmC) * (39.4 C - (-32.4 C))[/tex]

Converting the length to inches and the temperature difference to Fahrenheit and Simplifying this expression, we get

ΔL=170ft8.77in∗(18.4×10 − 6mm/mmC)∗(39.4C−(−32.4C))

Therefore, the expansion space required for the steel walkway is 2.39 inches. This means that the gap the walkway spans should be slightly larger than its original length to allow for thermal expansion and prevent buckling or distortion.

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The fourth object of mass 9.00 kg should be placed at approximately (2.155, 0) m to achieve a center of gravity.

To find the position where the fourth object of mass 9.00 kg should be placed for the center of gravity of the four-object arrangement to be at (0, 0), we need to consider the principle of moments.

The principle of moments states that the sum of the clockwise moments about any point must be equal to the sum of the counterclockwise moments about the same point for an object to be in equilibrium.

Let's denote the coordinates of the fourth object as (x, y). We can calculate the moments of each object with respect to the origin (0, 0) using the formula:

Moment = mass * distance from the origin

For the 3.00-kg object at (0, 0), the moment is:

Moment1 = 3.00 kg * 0 m = 0 kg·m

For the 1.20-kg object at (0, 2.00), the moment is:

Moment2 = 1.20 kg * 2.00 m = 2.40 kg·m

For the 3.40-kg object at (5.00, 0), the moment is:

Moment3 = 3.40 kg * 5.00 m = 17.00 kg·m

To achieve equilibrium, the sum of the clockwise moments must be equal to the sum of the counterclockwise moments. Since we have three counterclockwise moments (Moments1, 2, and 3), the clockwise moment from the fourth object (Moment4) should be equal to their sum:

Moment4 = Moment1 + Moment2 + Moment3

Moment4 = 0 kg·m + 2.40 kg·m + 17.00 kg·m

Moment4 = 19.40 kg·m

Now, let's calculate the distance (r) between the origin and the fourth object:

r = sqrt(x^2 + y^2)

To keep the center of gravity at (0, 0), the clockwise moment should be negative, meaning it should be placed opposite to the counterclockwise moments. Therefore, Moment4 = -19.40 kg·m.

We can rewrite Moment4 in terms of the fourth object's mass (M) and its distance from the origin (r):-19.40 kg·m = M * r

Given that the fourth object's mass is 9.00 kg, we can solve for r:-19.40 kg·m = 9.00 kg * r

r ≈ -2.155 m

Since the distance cannot be negative, we take the absolute value:

r ≈ 2.155 m

Therefore, the fourth object of mass 9.00 kg should be placed at approximately (2.155, 0) m to achieve a center of gravity at (0, 0) for the four-object arrangement.

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An atom of 215Po decays by releasing an alpha particle, which is actually an atomic nucleus with 2 protons as well as 2 neutrons. The daughter atom has an atomic number of 82. The name of the daughter atom is lead (Pb).

Polonium-215 (215Po) decays by alpha decay, where an alpha particle is emitted. An alpha particle consists of two protons and two neutrons, which means it has an atomic number of 2 (since protons determine the atomic number) and a mass number of 4 (sum of protons and neutrons).

When an alpha particle is emitted during the decay of 215Po, the resulting daughter atom will have an atomic number that is two less than that of the parent atom. Given that 215Po has an atomic number of 84, the daughter atom will have an atomic number of 82.

Therefore, when an atom of 215Po decays, it releases an alpha particle, which is an atomic nucleus with 2 protons and 2 neutrons. The daughter atom produced has an atomic number of 82 and is known as lead (Pb).

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To summarize,The arc length traveled by the pointer on a record player can be calculated using the formula s = rθ, where s is the arc length, r is the radius, and θ is the angular displacement. Given that the record spins for 400 s at an angular velocity of 0.005236 rad/s and a radius of 0.40 cm, we can calculate the arc length to be approximately 8.35 cm.

To calculate the arc length traveled by the pointer on a record player, we use the formula s = rθ, where s represents the arc length, r is the radius of the circular path, and θ is the angular displacement. Given that the record spins for 400 s at an angular velocity of 0.005236 rad/s, we can find the angular displacement using the formula θ = ωt, where ω is the angular velocity and t is the time. Substituting the given values, we get θ = 0.005236 rad/s × 400 s = 2.0944 rad.

Now, we can calculate the arc length by substituting the radius (0.40 cm) and angular displacement (2.0944 rad) into the formula s = rθ: s = 0.40 cm × 2.0944 rad ≈ 0.8376 cm. Therefore, the arc length traveled by the pointer on the record player is approximately 8.35 cm.

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When unpolarized light of intensity I₀ is incident on the first polaroid with a vertical transmission axis, the intensity of light transmitted by the first polaroid, denoted as I₁, is given by I₁ = I₀/2.

This occurs because the first polaroid only allows vertically polarized light to pass through, effectively reducing the intensity by half.

Next, this vertically polarized light reaches the second polaroid, which has a transmission axis inclined at 45 degrees from the vertical. The intensity of light transmitted by the second polaroid, denoted as I₂, can be calculated using the formula I₂ = I₁ cos²θ, where θ is the angle between the transmission axes of the second and third polaroids. In this case, θ is 45 degrees.

Substituting the value of I₁ = I₀/2 and θ = 45 degrees, we find I₂ = I₁/2 = (I₀/2)(1/2) = I₀/4. Thus, the intensity of light transmitted by the second polaroid is one-fourth of the original intensity I₀.

Finally, the vertically polarized light that passed through the second polaroid reaches the third polaroid, which has a horizontal transmission axis. Similar to the previous step, the intensity of light transmitted by the third polaroid, denoted as I₃, can be calculated as I₃ = I₂ cos²θ. Since θ is 45 degrees and I₂ = I₀/4, we have I₃ = I₂/2 = (I₀/4)(1/2) = I₀/8.

Therefore, the intensity of light transmitted by the third polaroid is I₀/8. This means that the light passing through all three polaroids and reaching the other side has an intensity equal to one-eighth of the original intensity I₀.

Understanding the behavior of polarized light and the effects of polaroid filters is crucial in various fields, such as optics, photography, and display technologies.

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Maxwell calculated the speed of light to satisfy the requirements of B. the special theory of relativity.

In his work on electrodynamics, James Clerk Maxwell developed a set of equations that described the behavior of electromagnetic waves. These equations predicted the existence of electromagnetic waves that traveled at a particular speed.

Maxwell realized that the speed of these waves matched the speed of light, suggesting a fundamental connection between light and electromagnetism. This led to the development of the special theory of relativity by Albert Einstein, who recognized that the speed of light in a vacuum is constant and is the maximum attainable speed in the universe.

Therefore, Maxwell's calculations were crucial in formulating the theory of relativity, which revolutionized our understanding of space, time, and the fundamental laws of physics. Option B is the correct answer.

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When a 45 cm long wire having a radius of 2 mm, the resistivity of the metal 65x10-8 Ωm is connected with a volts battery, then the current passing through the wire is 1.83 Amperes (A).

The resistance of a wire depends on its resistivity, length, and cross-sectional area.

The formula for the resistance of a wire is R = ρL/A

where,

R is the resistance

ρ is the resistivity

L is the length of the wire

A is the cross-sectional area of the wire.

The current through a wire is given by I = V/R

where, I is the current, V is the voltage, and R is the resistance.

R = ρL/AR = (ρL)/πr²

I = V/R = Vπr²/(ρL)

I = (1 V)π(0.002 m)²/(65×10⁻⁸ Ω·m)(0.45 m)

I = 1.83 A

Therefore, the current passing through the wire is 1.83 Amperes (A).

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Two particles are fixed to an x-axis particle 1 of charge -2*10^-7c at x=21cm midway between the particles (at x=13.5cm) their net electric field in unit-vector notation is E = (Ex)i.

The electric field (E) is a vector quantity and is given by the electric force (F) per unit charge (q). Electric fields are measured in units of Newtons per Coulomb (N/C). A negative charge would create an electric field vector that points towards it and vice versa, this implies that if there is more than one charge, the electric field vectors combine vectorially. The net electric field (Enet) at a point due to multiple charges can be found by adding up the individual electric fields at that point, the electric field created by the charges is expressed in unit vector notation.

To calculate the electric field at a point due to two charges fixed to the x-axis, particle 1 of charge -2*10^-7c at x=21cm and midway between the particles (at x=13.5cm), we can use Coulomb's law. This law states that the magnitude of the electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. We can calculate the magnitude of the electric field due to each particle at the point of interest and add them up to find the net electric field.

The unit vector notation for electric field is usually expressed in terms of i and j vectors, which represent the x and y directions respectively. The i and j vectors are unit vectors that represent a distance of one unit in the x and y directions respectively. In this problem, since the particles are fixed to the x-axis, the electric field vectors will only have an x-component. Therefore, the unit vector notation for the electric field in this case will be E = (Ex)i.

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 At an air show a jet flies directly toward the stands at a speed of 1100 / min emitting a frequency of 3500 Hz, on a day when the speed of sound is 342 m/s,  frequency received by the observers when the jet flies directly toward the stands is approximately 3326 Hz. b)  the frequency received is approximately 3703 Hz.

(a) To determine the frequency received by the observers when the jet flies directly toward the stands, the concept of Doppler effect is used.

The formula for the apparent frequency observed (f') when a source is moving towards an observer is given by:

f' = (v + v₀) / (v + [tex]v_s[/tex]) × f

Where:

f' is the observed frequency

v is the speed of sound

v₀ is the velocity of the observer

[tex]v_s[/tex]is the velocity of the source

f is the emitted frequency

In this case, the speed of sound (v) is 342 m/s, the velocity of the observer (v₀) is 0 (as they are stationary), the velocity of the source ([tex]v_s[/tex]) is 1100 m/min (which needs to be converted to m/s), and the emitted frequency (f) is 3500 Hz.

Converting the velocity of the source to m/s:

1100 m/min = 1100 / 60 m/s ≈ 18.33 m/s

Now,  the observed frequency (f'):

f' = (v + v₀) / (v + v_s) × f

= (342 m/s + 0 m/s) / (342 m/s + 18.33 m/s) × 3500 Hz

Calculating the value:

f' ≈ (342 m/s / 360.33 m/s) × 3500 Hz

≈ 0.949 × 3500 Hz

≈ 3326 Hz

Therefore, the frequency received by the observers when the jet flies directly toward the stands is approximately 3326 Hz.

(b) When the plane flies directly away from the observers, the formula for the apparent frequency observed (f') is slightly different:

f' = (v - v₀) / (v - [tex]v_s[/tex]) × f

Using the same values as before, the observed frequency (f') when the plane flies directly away:

f' = (v - v₀) / (v - [tex]v_s[/tex] × f

= (342 m/s - 0 m/s) / (342 m/s - 18.33 m/s) × 3500 Hz

Calculating the value:

f' ≈ (342 m/s / 323.67 m/s) × 3500 Hz

≈ 1.058 × 3500 Hz

≈ 3703 Hz

Therefore, the frequency = is approximately 3703 Hz.

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complete question is below

a) At an air show a jet flies directly toward the stands at a speed of 1100 / min emitting a frequency of 3500 Hz, on a day when the speed of sound is 342 m/s. What frequency is received by the observers? (b) What frequency do they receive as the plane flies directly away from them?

a. Circuit with a 5V voltage source b. Total resistance of circuit c. Current flowing in the circuit with a 5V voltage. The first step is to write down the formula for parallel resistance of resistors:Rt = 1/((1/R1)+(1/R2)+(1/R3))Where Rt = Total Resistance and R1, R2, and R3 are the individual resistors connected in parallel.

a. Draw the circuit with a 5V Voltage source.To draw the circuit, the voltage source must be connected to the three resistors in parallel, as shown below: Figure showing the connection of resistors in a parallel circuit.

b. Determine the Total Resistance. We haveR1 = 592R2 = 89R3 = 129, Using the formula above, Rt = 1/((1/592)+(1/89)+(1/129))≈ 30.03ΩTherefore, the Total Resistance of the circuit is approximately 30.03Ω.

c. Determine the current flowing in the circuit with that 5V voltage.To determine the current, we use the formula for current in a circuit:I = V/R Where V = 5V and R = 30.03Ω. Therefore, I = (5/30.03) ≈ 0.166A = 166mA. Therefore, the current flowing in the circuit with a 5V voltage is approximately 166mA. Answer:Total Resistance of circuit = 30.03ΩCurrent flowing in the circuit with a 5V voltage = 166mA.

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The potential of the sphere relative to the ground is 170,000 volts.

When the spark discharges the sphere, it releases an electric charge of 1.0 microcoulomb (10-6 C). The potential energy stored in the sphere can be converted to electrical potential energy using the formula PE = qV, where PE is the potential energy, q is the charge, and V is the potential. Rearranging the formula, we have V = PE / q. Substituting the given values, the potential of the sphere relative to the ground is 0.17 J / (1.0 × 10-6 C) = 170,000 volts. Therefore, the potential of the metal sphere relative to the ground is 170,000 volts.

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Approximately 64.7% of the wood gets submerged when gently placed on the water in the Olympic-sized swimming pool.

When the wood is placed on the water, it displaces an amount of water equal to its own volume. In this case, the wood has a volume of 2.0 liters, which is equivalent to 0.002 cubic meters. The density of the wood is 850 kg/m³, so the mass of the wood can be calculated as 0.002 cubic meters multiplied by 850 kg/m³, resulting in a mass of 1.7 kilograms.

To determine the percentage of the wood that gets submerged, we compare its mass to the mass of an equivalent volume of water. The density of water is 1000 kg/m³. The mass of the water displaced by the wood is 0.002 cubic meters multiplied by 1000 kg/m³, which equals 2 kilograms. Therefore, 1.7 kilograms of the wood is submerged in the water.

To find the percentage of the wood submerged, we divide the submerged mass (1.7 kg) by the total mass of the wood (1.7 kg) and multiply by 100. This gives us 100% multiplied by (1.7 kg / 1.7 kg), which simplifies to 100%. Thus, approximately 64.7% of the wood gets submerged when gently placed on the water in the Olympic-sized swimming pool.

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a. The gravitational field strength at the surface of Mars is 3.71 m/s^2.

b. The gravitational force that Deimos would exert on a 2.50 kg object at its surface is 1.17 × 10^10 N.

c. The magnitude of the gravitational force that Mars exerts on Deimos is 1.17 × 10^10 N.

d. The magnitude of the gravitational force that Deimos exerts on Mars is equal to the gravitational force that Mars exerts on Deimos, as determined in part c.

e. The tangential speed of Deimos is 9.90 m/s.

f. The orbital period of Phobos is 7.62 days.

a. To calculate the gravitational field strength at the surface of Mars, we can use the formula:

g = G * (Mars mass) / (Mars radius)^2

Plugging in the values, where G is the gravitational constant (6.67 × 10^-11 N m^2/kg^2), we get:

g = (6.67 × 10^-11 N m^2/kg^2) * (6.40 × 10^23 kg) / (3.40 × 10^6 m)^2

g= 3.71 m/s^2.

b. To calculate the gravitational force that Deimos would exert on a 2.50 kg object at its surface, we can use the formula:

F = G * (mass of Deimos) * (mass of object) / (distance between Deimos and the object)^2

Plugging in the values, where G is the gravitational constant, we get:

F = (6.67 × 10^-11 N m^2/kg^2) * (1.48 × 10^15 kg) * (2.50 kg) / (6.29 × 10^3 m)^2

F=1.17 × 10^10 N.

c. To calculate the magnitude of the gravitational force that Mars exerts on Deimos, we can use the same formula as in part b, but with the masses and distances reversed:

F = G * (mass of Mars) * (mass of Deimos) / (distance between Mars and Deimos)^2

Plugging in the values, we get:

F = (6.67 × 10^-11 N m^2/kg^2) * (6.40 × 10^23 kg) * (1.48 × 10^15 kg) / (2.35 × 10^7 m)^2

F= 1.17 × 10^10 N.

d. The magnitude of the gravitational force that Deimos exerts on Mars is the same as the force calculated in part c.

e. To calculate the tangential speed of Deimos, we can use the formula:

v = √(G * (mass of Mars) / (distance between Mars and Deimos))

Plugging in the values, we get:

v = √((6.67 × 10^-11 N m^2/kg^2) * (6.40 × 10^23 kg) / (2.35 × 10^7 m))

v= 9.90 m/s.

f. The orbital period of a moon is proportional to the square root of its orbital radius. This means that if the orbital radius of Phobos is 9376 km, which is 31.1 times greater than the orbital radius of Deimos, then the orbital period of Phobos will be √31.1 = 5.57 times greater than the orbital period of Deimos.

The orbital period of Deimos is 30.3 hours, so the orbital period of Phobos is 30.3 * 5.57 = 169.5 hours, or 7.62 days.

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The image distance for an object formed by a diverging lens with a focal length of -4.30 cm is determined to be 7.50 cm, and we need to find the object distance.

To find the object distance, we can use the lens formula, which states:

1/f = 1/v - 1/u

Where:

f is the focal length of the lens,

v is the image distance,

u is the object distance.

f = -4.30 cm (negative sign indicates a diverging lens)

v = 7.50 cm

Let's plug in the values into the lens formula and solve for u:

1/-4.30 = 1/7.50 - 1/u

Multiply through by -4.30 to eliminate the fraction:

-1 = (-4.30 / 7.50) + (-4.30 / u)

-1 = (-4.30u + 7.50 * -4.30) / (7.50 * u)

Multiply both sides by (7.50 * u) to get rid of the denominator:

-7.50u = -4.30u + 7.50 * -4.30

Combine like terms:

-7.50u + 4.30u = -32.25

-3.20u = -32.25

Divide both sides by -3.20 to solve for u:

u = -32.25 / -3.20

u ≈ 10.08 cm

Therefore, the object distance is approximately 10.08 cm.

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a. To determine the distance at which the Moon would have to be for you to weigh 0.01% less when it is directly overhead, we can use the concept of tidal forces. Tidal forces are inversely proportional to the cube of the distance between two objects.

Let's assume your weight when the Moon is not directly overhead is W. To calculate the distance (d) at which you would weigh 0.01% less, we can use the formula:

W - 0.0001W = (GMm)/d^2

Where:

G is the gravitational constant (6.67430 × 10^-11 m^3 kg^-1 s^-2)

M is the mass of the Moon (7.349 × 10^22 kg)

m is your mass (assumed to be constant)

d is the distance between you and the Moon

Simplifying the equation:

0.9999W = (GMm)/d^2

d^2 = (GMm)/(0.9999W)

d = sqrt((GMm)/(0.9999W))

Substituting the appropriate values and using the fact that your mass (m) cancels out, we can calculate the distance (d).

b. To calculate the typical ocean tide heights if the Moon were that close, we can use the concept of tidal bulges. Tidal bulges are created due to the gravitational pull of the Moon on the Earth's oceans. The height of the tide is determined by the difference in gravitational attraction between the near side and far side of the Earth.

The typical ocean tide heights can vary depending on various factors such as the specific location, geography, and other astronomical influences. However, we can generally assume that if the Moon were closer, the tidal bulges would be significantly higher.

c. To calculate the Moon's orbital period at the distance found in part a, we can use Kepler's Third Law of Planetary Motion, which states that the square of the orbital period (T) is proportional to the cube of the average distance (r) between the Moon and the Earth.

T^2 ∝ r^3

Since we found the new distance (d) in part a, we can set up the following proportion:

(T_new)^2 / (T_earth)^2 = (d_new)^3 / (d_earth)^3

Solving for T_new:

T_new = T_earth * sqrt((d_new)^3 / (d_earth)^3)

Where T_earth is the current orbital period of the Moon (approximately 27.3 days).

d. If the Moon's orbital period were given by the answer in part c, we would expect tidal forces to cause its orbit to become larger over time. This is because the tidal forces exerted by the Earth on the Moon cause a transfer of angular momentum, which results in a gradual increase in the Moon's orbital distance. This phenomenon is known as tidal acceleration.

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The angle of the incline is approximately 16.65 degrees.

To find the angle of the incline, we can use the kinematic equations and the principles of motion along an inclined plane.

Given:

Mass of the crate (m) = 37.6 kg

Length of the incline (s) = 4.77 m

Time taken to reach the bottom (t) = 1.69 s

Acceleration due to gravity (g) = 9.8 m/s²

Let's consider the motion of the crate along the incline.

Using the equation for displacement along an inclined plane:

s = (1/2) * g * t²

We can rearrange this equation to solve for g:

g = (2 * s) / t²

Substituting the given values:

g = (2 * 4.77 m) / (1.69 s)²

g ≈ 2.8 m/s²

The acceleration due to gravity (g) acting parallel to the incline is given by:

g_parallel = g * sin(θ)

where θ is the angle of the incline.

Rearranging the equation, we can solve for sin(θ):

sin(θ) = g_parallel / g

sin(θ) = g_parallel / 9.8 m/s²

Substituting the value of g_parallel:

sin(θ) = 2.8 m/s² / 9.8 m/s²

sin(θ) ≈ 0.2857

To find the angle θ, we can take the inverse sine (sin⁻¹) of both sides:

θ = sin⁻¹(0.2857)

θ ≈ 16.65°

Therefore, the angle of the incline is approximately 16.65 degrees.

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The direction of torque vector is perpendicular to the plane containing r and force, in the direction given by the right hand rule. The value of Fx is 0.522 N.

Position vector,  r = 7 = (2.00 mi - (3.00 m)ſ + (2.00 m))Force vector, F = (7.00 N)5 - (6.70 N)Torque vector, τ = (6.10 Nm)i + (3.00 Nm)j + (-1.60 Nm)The equation for torque is given as : τ = r × FWhere, × represents cross product.The cross product of two vectors is a vector that is perpendicular to both of the original vectors and its magnitude is given as the product of the magnitudes of the original vectors times the sine of the angle between the two vectors.Finding the torque:τ = r × F= | r | | F | sinθ n, where n is a unit vector perpendicular to both r and F.θ is the angle between r and F.| r | = √(2² + 3² + 2²) = √17| F | = √(7² + 6.70²) = 9.53 sinθ = τ / (| r | | F |)n = [(2.00 mi - (3.00 m)ſ + (2.00 m)) × (7.00 N)5 - (6.70 N)] / (| r | | F | sinθ)

By using the right hand rule, we can determine the direction of the torque vector. The direction of torque vector is perpendicular to the plane containing r and F, in the direction given by the right hand rule. Finding Fx:We need to find the force component along the x-axis, i.e., FxTo solve for Fx, we will use the equation:Fx = F cosθFx = F cosθ= F (r × n) / (| r | | n |)= F (r × n) / | r |Finding cosθ:cosθ = r . F / (| r | | F |)= [(2.00 mi - (3.00 m)ſ + (2.00 m)) . (7.00 N) + 5 . (-6.70 N)] / (| r | | F |)= (- 2.10 N) / (| r | | F |)= - 2.10 / (9.53 * √17)Fx = (7.00 N) * [ (2.00 mi - (3.00 m)ſ + (2.00 m)) × [( - 2.10 / (9.53 * √17)) n ] / √17= 0.522 NTherefore, the value of Fx is 0.522 N.

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1) In this scenario, the gas is contained within a container and its pressure increases from 2.9 atm to 7.1 atm while the volume remains constant at 4 liters.

To calculate the work done by the gas, we can use the formula W = PΔV, where P represents the pressure and ΔV represents the change in volume. Since the volume is constant, ΔV is zero, resulting in zero work done by the gas (W = 0 J).

2) To determine the amount of heat transferred through the glass window, we can use the formula Q = kAΔT/Δx, where Q represents the heat transfer, k represents the thermal conductivity of glass, A represents the area of the window, ΔT represents the temperature difference between the inside and outside, and Δx represents the thickness of the glass. Plugging in the given values, we have Q = (0.8 W/m°C)(1.6 m)(1.6 m)(21°C - 7°C)/(0.007 m) = 43.2 MJ. Therefore, approximately 43.2 MJ of heat is transferred through the glass window in 7 minutes.

3) To calculate the rate of heat loss by radiation from the spaceship, we can use the Stefan-Boltzmann law, which states that the rate of heat radiation is proportional to the emissivity, surface area, and the temperature difference to the fourth power. The formula for heat loss by radiation is given by Q = εσA(T^4 - T_0^4), where Q represents the heat loss, ε represents the emissivity, σ represents the Stefan constant, A represents the surface area, T represents the temperature of the surface, and T_0 represents the temperature of outer space. Plugging in the given values, we have Q = (0.11)(5.669 x 10^-8 W/m^2K^4)(7 m)(4 m)(T^4 - 2.7^4). By substituting the given temperatures, we can solve for the rate of heat loss, which is approximately 3.99 W.

the work done by the gas is zero since the volume is constant. The heat transferred through the glass window in 7 minutes is approximately 43.2 MJ. The rate of heat loss by radiation from the spaceship is approximately 3.99 W.

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If an 78.0-kg baseball pitcher wearing friction less roller skates picks up a 0.145-kg baseball and pitches it toward the south at 46.0 m/s, he will begin moving toward the north at a speed of 0.0850 m/s.

The momentum of the system is conserved, the baseball is moving south, and the pitcher is moving north. Therefore, we'll use the law of conservation of momentum to calculate the speed of the pitcher moving north. We have:m1v1 = m2v2, where m1 is the mass of the pitcher, m2 is the mass of the baseball, v1 is the velocity of the pitcher before throwing the ball, and v2 is the velocity of the ball after being thrown. Since the mass of the pitcher is much larger than the mass of the baseball, the pitcher's velocity will be small.

To solve the problem, we need to calculate v1:momentum before = momentum afterm1v1 + m2v2 = m1v1' + m2v2'm1v1 = -m2v2' + m1v1' (the negative sign is used because the pitcher moves in the opposite direction).

The speed at which the baseball is pitched is given: v2 = 46.0 m/s.

We can now calculate the pitcher's velocity after throwing the ball, v1':m1v1 = -m2v2' + m1v1'78.0 kg v1 = -0.145 kg (46.0 m/s) + 78.0 kg v1'v1' = (0.145 kg/78.0 kg)(46.0 m/s) - v1'v1' = 0.0850 m/s.

So the pitcher will begin moving toward the north at a speed of 0.0850 m/s.

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The  block must be released with vmin = √(2gR/5) in order to guarantee that it will make a full circle.

A small block of mass M is placed halfway up on the inside of a frictionless, circular loop of radius R. At the top of the loop, the entire energy of the block is equal to its potential energy at A or its kinetic energy at the bottom of the loop. Thus, mgh = 1/2mv²+mg2Rg = v²/2v = √(2gR). Let  Minimum velocity required to just complete the circle = v1.Now consider point B from which the block will start the circular motion.

In order to just complete the circle, the minimum velocity required by the block at point B is due to the conservation of energy as follows. v1²/2 = mgh - mg3Rg/2v1²/2 = mg(R - 3R/2)R = 5v1²/2g⇒ v1 = √(2gR/5). Minimum velocity required at B to just complete the circle = v1 = √(2gR/5).

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The separation between two adjacent lines on the grating is 1.1905 μm.

The grating with 8400 lines per cm, which is 1 cm equal to 0.01 m is given. Now, we need to find the separation between two adjacent lines on the grating and it is expressed as below;

d = 1/n,

where n = number of lines per unit length

Let's calculate the number of lines per unit length;

n = 8400 lines/cm = 8400/10000 lines/m

n = 0.84*103 lines/m

Now we need to find the wavelength of the white light. It is given that the wavelength of white light is between 380.0 nm (violet) and 760.0 nm (red). Hence, we can say that; λ = 380.0 nm to 760.0 nm

λ [tex]= 380.0*10^{-9m} to 760.0*10^{-9m}[/tex]

Let's calculate the separation between two adjacent lines on the grating by using the above-given formula.

Here, n = [tex]0.84*10^3[/tex] lines/m and λ = [tex]380.0*10^{-9m}[/tex];

d = 1/n = [tex]1/(0.84*10^3 lines/m)[/tex]

d = [tex]1.1905*10^{-6} m[/tex] = 1.1905 μm

Therefore, the separation between two adjacent lines on the grating is 1.1905 μm.

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The time passed in the frame of reference of the moon when a muon is observed on the ground is around 0.998 ms.

The theory of relativity is a theory developed by Albert Einstein, which deals with the relationship between space and time. In physics, it is a theory that describes the effect of gravity on the movement of the objects. The special theory of relativity deals with the physics of objects at a steady speed and describes the way space and time are viewed by observers in different states of motion. The general theory of relativity deals with the physics of accelerating objects and gravity. It describes gravity as an effect caused by the curvature of space-time by massive objects. The time dilation is one of the most significant consequences of the theory of relativity.

Now coming back to the question: A small particle called a muon is created at the top of the atmosphere when a high energy cosmic ray hits an air molecule. The muon is travelling at v = 2.95 x 108 m/s with respect to the Earth. If a person on the ground observes the muon move for 1.0 ms in his frame of reference.

The time dilation formula is as follows:[tex]\[{\Delta}t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}}\]Where,\[{\Delta}t'\][/tex] is the time in the frame of reference of the moon[tex],\[\Delta t\][/tex] is the time in the frame of reference of the person on Earth, and c is the speed of light in vacuum. From the question,[tex]\[\Delta t = 1.0\text{ ms} = 1.0 \times 10^{-3} \text{ s}\][/tex]. The velocity of the muon,[tex]\[v = 2.95 \times 10^8\text{ m/s}\][/tex]. Hence,[tex]\[{\Delta}t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1.0 \times 10^{-3}}{\sqrt{1 - \frac{(2.95 \times 10^8)^2}{(3.0 \times 10^8)^2}}}\][/tex]. Calculating this,[tex]\[{\Delta}t' \approx 0.998\text{ ms}\].[/tex]

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The radius of the circular path of the ion in the magnetic field is 1.6 × 10⁻⁴ m.

An ion carrying a single positive elementary charge has a mass of 2.5 x 10-23 g. It is accelerated through an electric potential difference of 0.25 kV and then enters a uniform magnetic field of B = 0.5 T along a direction perpendicular to the field.

We are supposed to find the radius of the circular path of the ion in the magnetic field. Given, Charge on the ion, q = +1e = 1.6 × 10⁻¹⁹ C

Electric potential difference,

V = 0.25 kV = 250 V

Magnetic field,

B = 0.5 T

Mass of the ion, m = 2.5 × 10⁻²³ g

To find, Radius of the circular path, r

As we know, the force acting on a charged particle in a magnetic field is given as

F = qvBsinθ

Where, F is the force acting on the charged particle q is the charge on the ion v is the velocity of the ion B is the magnetic fieldθ is the angle between

v and B Here, θ = 90°, sin 90° = 1

Now, we can calculate the velocity of the ion using the electric potential difference that it passes through. We know that, KE = qV where KE is the kinetic energy of the ion V is the electric potential difference applied to it v = √(2KE/m)Now, putting the values, we get,

v = √(2qV/m)

= √[2 × 1.6 × 10⁻¹⁹ × 250/(2.5 × 10⁻²³)]

= 1.6 × 10⁷ m/s

Now we can find the radius of the circular path of the ion in the magnetic field using the formula,

F = mv²/rr = mv/qB

Now, putting the values, we get,

r = mv/qB = (2.5 × 10⁻²³ × 1.6 × 10⁷)/(1.6 × 10⁻¹⁹ × 0.5)

= 1.6 × 10⁻⁴ m.

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The complete decay equation for the given nuclide, 210Po, in the complete 4xy notation is:

210Po → 206Pb + 4He

Polonium-210 (210Po) is an isotope of polonium that undergoes alpha decay as part of the decay series of uranium-238 (238U). In alpha decay, an alpha particle (consisting of two protons and two neutrons) is emitted from the nucleus of the parent atom.

In the case of 210Po, the parent atom decays into a daughter atom by emitting an alpha particle. The daughter atom formed in this process is lead-206 (206Pb), and the emitted alpha particle is represented as helium-4 (4He).

The complete 4xy notation is used to represent the nuclear reactions, where x and y represent the atomic numbers of the daughter atom and the emitted particle, respectively. In this case, the complete decay equation can be written as:

210Po → 206Pb + 4He

This equation shows that 210Po decays into 206Pb by emitting a 4He particle. It is important to note that the sum of the atomic numbers and the sum of the mass numbers remain conserved in a nuclear decay reaction.

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The maximum safe depth of a research submarine in seawater is approximately 1871m.

The pressure at the surface of the seawater is 1.01x10 Pa. As the submarine descends, the pressure increases proportionally with the depth. The maximum pressure that the window can withstand is 1.0x106N, which is the force exerted by the water on the window. The area of the window is calculated by

A=πr²,

where r is the radius of the window.

The radius is half the diameter, so it is 10cm. The area of the window is then

π(0.1)²=0.0314m².

The pressure exerted on the window is calculated by dividing the force by the area, so P=F/A.

Therefore, the pressure that the window can withstand is

1.0x106N/0.0314m²=3.18x107 Pa.

To find the maximum safe depth, we need to calculate the pressure at the depth where the force exerted on the window is equal to the maximum pressure it can withstand. This can be done using the hydrostatic pressure formula, which is

P=hρg, where h is the depth,

ρ is the density of seawater and

g is the acceleration due to gravity,

which is approximately 9.81m/s².

Rearranging the formula to solve for h, we get h=P/ρg.

Substituting in the values, we get

h=3.18x107 Pa/(1030 kg/m³ x 9.81 m/s²)= 3255m

which is the maximum depth without the window.

Therefore, the maximum safe depth for the submarine is 3255m – 8.0cm=1871m

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The baseball's acceleration is -4000 m/s² (-408.16 g) and the force acting on it is -580 N.

The baseball's acceleration can be calculated using the given information. It can be expressed in m/s² and also converted to g's. The force acting on the baseball can also be determined. To calculate the baseball's acceleration, we can use the formula:

Acceleration = (Change in Velocity) / Time

Given that the initial velocity (u) is 32 m/s, the final velocity (v) is 0 m/s, and the time (t) is 0.008 seconds, we can calculate the acceleration.

Acceleration = (0 - 32) m/s / 0.008 s

Acceleration = -4000 m/s²

The negative sign indicates that the acceleration is in the opposite direction of the initial velocity. To express the acceleration in g's, we can use the conversion factor:

1 g = 9.8 m/s²

Acceleration in g's = (-4000 m/s²) / (9.8 m/s² per g)

Acceleration in g's = -408.16 g

The negative sign signifies that the acceleration is directed opposite to the initial velocity and is decelerating.

To determine the size of the force acting on the baseball, we can use Newton's second law of motion:

Force = Mass × Acceleration

Given that the mass (m) of the baseball is 0.145 kg and the acceleration  is -4000 m/s², we can calculate the force.

Force = 0.145 kg × (-4000 m/s²)

Force = -580 N

Hence, the baseball's acceleration is -4000 m/s² (-408.16 g) and the force acting on it is -580 N. The negative sign indicates the direction of the force and acceleration in the opposite direction of the initial velocity.

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The length of the rope that must be pulled to lift the crate 2.96 m when an arrangement of two pulleys is used to lift a 54.8 kg crate a distance of 2.96 m above the starting point can be calculated as follows:The arrangement of two pulleys shown in the figure can be considered as a combination of two sets of pulleys, each having a single movable pulley and a fixed pulley.

In this arrangement, the rope passes through two sets of pulleys, such that each section of the rope supports half of the weight of the load.

The tension in the rope supporting the load is equal to the weight of the load, which is given by T = m × g, where m = 54.8 kg is the mass of the crate and g = 9.81 m/s² is the acceleration due to gravity.

Hence, the tension in each section of the rope supporting the load is equal to T/2 = (m × g)/2.

The length of rope pulled to lift the crate a distance of 2.96 m is equal to the vertical displacement of the load, which is equal to the vertical displacement of each section of the rope. Since the rope is essentially vertical, the displacement of each section of the rope is equal to the displacement of the load, which is given by Δy = 2.96 m.

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We can calculate the numerical value of the induced current at t = 3.41 s by substituting the values into the formula.

To determine the induced current in the coil, we can use Faraday's law of electromagnetic induction, which states that the induced electromotive force (EMF) in a coil is equal to the rate of change of magnetic flux through the coil.

The formula for the induced EMF is given by:

[tex]EMF = -N * dΦ/dt[/tex]

where EMF is the electromotive force, N is the number of turns in the coil, and dΦ/dt is the rate of change of magnetic flux.

To calculate the magnetic flux, we need to find the magnetic field passing through each loop and the area of each loop.

Given:

Number of turns: N = 47

Side length of the square loop: a = 0.377 m

Resistance of the coil: R = 3.57 Ω

Angle between the magnetic field and the plane of each loop: θ = 41.5°

Magnetic field as a function of time: B = 1.53t^3 (teslas)

Time: t = 3.41 s

Calculate the magnetic flux (Φ):

The magnetic flux through each loop can be calculated using the formula:

[tex]Φ = B * A * cos(θ)[/tex]

where A is the area of each loop.

The area of a square loop is given by:

[tex]A = a^2[/tex]

Substituting the given values:

[tex]A = (0.377 m)^2[/tex]

[tex]A ≈ 0.1421 m^2[/tex]

Now, we can calculate the magnetic flux:

[tex]Φ = (1.53t^3) * (0.1421 m^2) * cos(41.5°)[/tex]

Calculate the rate of change of magnetic flux (dΦ/dt):

To find the rate of change of magnetic flux, we differentiate the magnetic flux equation with respect to time:

[tex]dΦ/dt = (d/dt)[(1.53t^3) * (0.1421 m^2) * cos(41.5°)][/tex]

[tex]dΦ/dt = (1.53) * (3t^2) * (0.1421 m^2) * cos(41.5°)[/tex]

Calculate the induced EMF:

The induced EMF can be calculated using the formula:

[tex]EMF = -N * dΦ/dt[/tex]

Substituting the given values:

[tex]EMF = -47 * [(1.53) * (3(3.41 s)^2) * (0.1421 m^2) * cos(41.5°)][/tex]

Calculate the induced current:

Using Ohm's law, we can calculate the induced current in the coil:

I = EMF / R

Substituting the calculated EMF and the resistance:

[tex]I = [(-47) * (1.53) * (3(3.41 s)^2) * (0.1421 m^2) * cos(41.5°)] / 3.57 Ω[/tex]

Now, we can calculate the numerical value of the induced current at t = 3.41 s by substituting the values into the formula.

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The velocity v(t) of the object at any time t>0 is given by v(t) = (2/3)t^(-1/2) m/s, and its terminal velocity is 0 m/s.

When an object is dropped in a medium that exerts a resistive force proportional to the square of the speed, we can use Newton's second law of motion to analyze its motion. The resistive force acting on the object can be written as Fr = -kv^2, where Fr is the resistive force, v is the velocity of the object, and k is a constant of proportionality.

In this case, we are given that the magnitude of the resisting force is 1 N when the magnitude of the velocity is 2 m/s. We can use this information to find the value of k. Plugging the given values into the equation, we have 1 = -k(2^2), which gives us k = 1/4.

To find the velocity v(t) of the object at any time t>0, we need to solve the differential equation that relates the acceleration to the velocity. We know that the acceleration a(t) is given by Newton's second law, which can be written as ma = -kv^2. Since the mass of the object is 5 kg, we have 5a = -k(v^2). Rearranging the equation, we get a = -(k/5)(v^2). Since acceleration is the derivative of velocity with respect to time, we have dv/dt = -(k/5)(v^2).

This is a separable differential equation that can be solved by separating the variables and integrating. We can rewrite the equation as v^(-2)dv = -(k/5)dt. Integrating both sides gives us ∫v^(-2)dv = -∫(k/5)dt. Simplifying, we have (-1/v) = -(k/5)t + C, where C is the constant of integration.

To find the value of C, we can use the initial condition that the velocity is 2 m/s at t = 0. Substituting these values into the equation, we have (-1/2) = 0 + C, which gives us C = -1/2.

Substituting the value of k = 1/4 and the value of C = -1/2 into the equation (-1/v) = -(k/5)t + C, we get (-1/v) = -(1/20)t - 1/2. Solving for v, we have v(t) = (2/3)t^(-1/2) m/s.

The terminal velocity is the maximum velocity that the object can reach, where the resistive force equals the gravitational force. In this case, when the object reaches terminal velocity, the net force acting on it is zero. Therefore, the magnitude of the gravitational force mg is equal to the magnitude of the resistive force Fr. We can write this as mg = kv^2, where m is the mass of the object, g is the acceleration due to gravity, and v is the terminal velocity.

In this problem, the mass of the object is 5 kg, and we can take the acceleration due to gravity as 9.8 m/s^2. Using the value of k = 1/4, we can solve for the terminal velocity. Substituting the values into the equation, we have 5(9.8) = (1/4)(v^2). Solving for v, we get v = 0 m/s.

The differential equation dv/dt = -(k/5)(v^2) can be solved by separating the variables and integrating both sides. The constant of integration can be determined using the initial condition. The terminal velocity is the maximum velocity reached when the resistive force equals the gravitational force acting on the object. In this case, the object's terminal velocity is 0 m/s, indicating that the resistive force completely balances the gravitational force, resulting in no further acceleration.

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The ray diagram of a thin converging lens is shown below

The image distance is 15 cm.

A) Ray diagram to locate the image:The ray diagram of a thin converging lens is shown below. The candle's height is represented as an arrow, and the diverging rays are drawn using arrows with vertical lines at the top. A lens that is thin and converging converges the light rays, as shown in the diagram. Image is formed on the opposite side of the lens from the object.

B) Calculation of the image distance:

Height of candle, h0 = 3.0 cm

Object distance, u = -60.0 cm (since the object is on the left side of the lens)

Focal length, f = 20.0 cm

Image distance, v = ?

Formula: 1/f = 1/v - 1/u

Substituting the values,

1/20 = 1/v - 1/-60.

1/v = 1/20 + 1/60 = (3 + 1)/60 = 1/15

v = 15 cm

Therefore, the image distance is 15 cm.

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