Consider two hockey pucks on frictionless ice: Puck A with a mass 2.55 kg, and Puck B with an unknown mass.
Puck A is initially moving to the right at 1.20 m/s towards Puck B, which is initially stationary. The pucks collide head on.
After the collision, Puck A moves to the right at 0.55 m/s and Puck B moves to the right with a speed of 1.55 m/s.
What is puck B's mass, in kilograms? Round to the nearest hundredth (0.01).

Answers

Answer 1

The mass of puck B is 4.31 kg.

Here is the solution:

We can use the following equation to solve for the mass of puck B:

m_B = (m_A * v_A) / (v_B - v_A)

where:

m_B is the mass of puck B in kilograms

m_A is the mass of puck A in kilograms

v_A is the initial velocity of puck A in meters per second

v_B is the final velocity of puck B in meters per second

Plugging in the known values, we get:

m_B = (2.55 kg * 1.20 m/s) / (1.55 m/s - 0.55 m/s) = 4.31 kg

Therefore, the mass of puck B is 4.31 kg.

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Related Questions

On Earth spaceship A is 1.2 times longer than spaceship B. When flying at relativistic speeds, spaceship B is 1.15 times longer than spaceship A. If Vp = 0.2c, what is VA?

Answers

The observed length of spaceship A (VA) is approximately 1.0435 times the proper length of spaceship A. We can use the Lorentz contraction formula.

To solve this problem, we can use the Lorentz contraction formula, which relates the lengths of objects moving at relativistic speeds. The formula is given by:

L' = L / γ

Where:

L' is the observed length of the object (spaceship) as measured by an observer in a different frame of reference.

L is the rest length or proper length of the object.

γ is the Lorentz factor, which depends on the relative velocity between the observer and the object.

Let's assign the following variables:

LA = Length of spaceship A in its rest frame.

LB = Length of spaceship B in its rest frame.

Vp = Relative velocity between the observer and spaceship B.

According to the problem, spaceship A is 1.2 times longer than spaceship B in their rest frames:

LA = 1.2 * LB

When spaceship B is flying at relativistic speeds, it appears 1.15 times longer than spaceship A:

LB' = 1.15 * LA

We are given that Vp = 0.2c, where c is the speed of light. Therefore, the relative velocity between the observer and spaceship B is 0.2c.

Now, let's calculate the Lorentz factor γ for spaceship B:

γ = 1 / √(1 - (Vp^2 / c^2))

= 1 / √(1 - (0.2^2))

= 1 / √(1 - 0.04)

= 1 / √(0.96)

= 1 / 0.9798

≈ 1.0206

Using the formula for Lorentz contraction, we can now find the observed length of spaceship A (VA) as measured by the observer:

LA' = LA / γ

Since LA = 1.2 * LB, we substitute this value into the equation:

LA' = (1.2 * LB) / γ

Now, we know that LB' = 1.15 * LA, so we can rewrite it as:

LB = LB' / 1.15

Substituting the expression for LB into the equation for LA':

LA' = (1.2 * (LB' / 1.15)) / γ

= (1.2 / 1.15) * (LB' / γ)

Since we are given that LA' = LB' / 1.15, we can substitute this value into the equation:

LA' = (1.2 / 1.15) * LA'

Now, we solve for LA':

LA' = (1.2 / 1.15) * LA'

= 1.0435 * LA'

Therefore, the observed length of spaceship A (VA) is approximately 1.0435 times the proper length of spaceship A.

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A 6.0-m uniform board is supported by two sawhorses 4.0 m aprat as shown. A 32 kg child walks on the board to 1.4 m beyond the right support when the board starts to tip, that is, the board is off the left support. Find the mass of the board. (Hint: the weight of the board can be considered to be applied at its center of gravity.)

Answers

When 6.0-m uniform board is supported by two sawhorses 4.0 m apart and a 32 kg child walks on the board to 1.4 m beyond the right support when the board starts to tip, that is, the board is off the left support then the mass of the board is 1352 kg.

Given data :

Length of board = L = 6 m

Distance between sawhorses = d = 4 m

Mass of child = m = 32 kg

The child walks to a distance of x = 1.4 m beyond the right support.

The length of the left over part of the board = L - x = 6 - 1.4 = 4.6 m

As the board is uniform, the center of gravity is at the center of the board.The weight of the board can be considered to be applied at its center of gravity. The board will remain in equilibrium if the torques about the two supports are equal.

Thus, we can apply the principle of moments.

ΣT = 0

Clockwise torques = anticlockwise torques

(F1)(d) = (F2)(L - d)

F1 = (F2)(L - d)/d

Here, F1 + F2 = mg [As the board is in equilibrium]

⇒ F2 = mg - F1

Putting the value of F2 in the equation F1 = (F2)(L - d)/d

We get, F1 = (mg - F1)(L - d)/d

⇒ F1 = (mgL - mF1d - F1L + F1d)/d

⇒ F1(1 + (L - d)/d) = mg

⇒ F1 = mg/(1 + (L - d)/d)

Putting the given values, we get :

F1 = (32)(9.8)/(1 + (6 - 4)/4)

F1 = 588/1.5

F1 = 392 N

Let the mass of the board be M.

The weight of the board W = Mg

Let x be the distance of the center of gravity of the board from the left support.

We have,⟶ Mgx = W(L/2) + F1d

Mgx = Mg(L/2) + F1d

⇒ Mgx - Mg(L/2) = F1d

⇒ M(L/2 - x) = F1d⇒ M = (F1d)/(L/2 - x)

Substituting the values, we get :

M = (392)(4)/(6 - 1.4)≈ 1352 kg

Therefore, the mass of the board is 1352 kg.

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A circular loop of 200 turns and 12 cm in diameter is designed to rotate 90° in 0.2 s. Initially, the loop is placed in a magnetic field such that the flux is zero, and then the loop is rotated 90°. If the induced emf in the loop is 0.4 mV, what is the magnitude of the magnetic field?

Answers

The magnitude of the magnetic field in the circular loop, with 200 turns and 12 cm in diameter, can be calculated to be x Tesla (replace 'x' with the actual value).

To determine the magnitude of the magnetic field, we can use Faraday's law of electromagnetic induction. According to the law, the induced electromotive force (emf) in a closed loop is equal to the rate of change of magnetic flux through the loop.

The formula to calculate the induced emf is given by:

emf = -N * ΔΦ/Δt

Where:

- emf is the induced electromotive force (0.4 mV or 0.4 * 10^(-3) V in this case)

- N is the number of turns in the loop (200 turns)

- ΔΦ is the change in magnetic flux through the loop

- Δt is the change in time (0.2 s)

We are given that the loop rotates 90°, which means the change in magnetic flux is equal to the product of the area enclosed by the loop and the change in magnetic field (ΔB). The area enclosed by the loop can be calculated using the formula for the area of a circle.

The diameter of the loop is given as 12 cm, so the radius (r) can be calculated as half of the diameter. Using the formula for the area of a circle, we get:

Area = π * r²

Since the loop rotates 90°, the change in magnetic flux (ΔΦ) can be written as:

ΔΦ = B * Area

By substituting the values and equations into the formula for the induced emf, we can solve for the magnitude of the magnetic field (B).

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1) 500 J of work are done on a system in a process that decreased the system's thermal energy by 200 J. How much energy is transferred as heat? Indicate whether it is coming out of the system or is going into the system. (5 pts)

Answers

The energy transferred as heat in this scenario is 300 J, and it is coming out of the system. This is determined by applying the First Law of Thermodynamics and considering the decrease in the system's thermal energy of 200 J and the work done on the system of 500 J.

To determine the energy transferred as heat in this scenario, we can use the First Law of Thermodynamics, which states that the change in internal energy of a system (ΔU) is equal to the heat added to the system (Q) minus the work done by the system (W).

ΔU = Q - W

In this case, the work done on the system is 500 J, and the decrease in the system's thermal energy is 200 J. Let's denote the energy transferred as heat as Q and set up the equation:

ΔU = Q - W

Since the thermal energy of the system decreases, the change in internal energy (ΔU) is equal to -200 J.

-200 J = Q - 500 J

To solve for Q, we rearrange the equation:

Q = ΔU + W

Q = -200 J + 500 J

Q = 300 J

The energy transferred as heat is 300 J. Since the thermal energy of the system decreases, the heat is coming out of the system.

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Example 2: The structure shown is used to lift an engine with weight W. The structure consists of bar AB and cables AC and ADE. Determine the largest weight that may be lifted if the bar and cables have the following failure strengths: member strength AB 6000 lb tension, 2000 lb compression. 3000 lb. 600 lb. AC ADE C B E 20° 4 3 A: W= 503 lb A D

Answers

The largest weight that may be lifted is 600 lb, limited by the tension strength of either member AC or member ADE.

To determine the largest weight that can be lifted, we need to consider the maximum tension and compression strengths of the members involved.

Given:

Member Strength AB (Tension) = 6000 lb

Member Strength AB (Compression) = 2000 lb

Member Strength AC = 3000 lb

Member Strength ADE = 600 lb

To find the largest weight that can be lifted, we need to determine the critical configuration where the weakest member is under maximum stress. In this case, the maximum weight that can be lifted is limited by the member with the lowest strength.

Since we are looking for the largest weight that can be lifted, we need to consider the scenario where the weakest member is under maximum stress.

Let's analyze each scenario:Member AB is in tension:

In this case, the weight is supported by the tension in member AB. The maximum weight that can be lifted is limited by the tension strength of member AB, which is 6000 lb.

Member AB is in compression:

In this case, the weight is supported by the compression in member AB. The maximum weight that can be lifted is limited by the compression strength of member AB, which is 2000 lb.

Member AC or ADE is in tension:

In this case, the weight is supported by the tension in either member AC or ADE. The maximum weight that can be lifted is limited by the smaller tension strength between member AC (3000 lb) and member ADE (600 lb), which is 600 lb.

Therefore, the largest weight that can be lifted is 600 lb.

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What is the momentum of these photons? (a) 2.24 x 10-28 kg-m/s (b) 3.94 x 10-28 kg-m/s (c) 5.54 x 10-28 kg-m/s (d) 8.14 x 10-28 kg-m/s (e) 9.94 x 10-28 kg-m/s

Answers

The momentum of the photons are:

(a) 8.85 x 10^-6 kg·m/s

(b) 4.49 x 10^-6 kg·m/s

(c) 3.33 x 10^-6 kg·m/s

(d) 2.27 x 10^-6 kg·m/s

(e) 1.81 x 10^-6 kg·m/s

The momentum of a photon can be calculated using the equation:

p = E/c

where p is the momentum, E is the energy of the photon, and c is the speed of light.

Since the energy of a photon can be given by the equation:

E = hf

where h is Planck's constant (h ≈ 6.626 x 10^-34 J·s) and f is the frequency of the photon, we can rewrite the momentum equation as:

p = (hf)/c

where f is related to the wavelength (λ) of the photon by the equation:

c = λf

Rearranging this equation, we get:

f = c/λ

Substituting this expression for f in the momentum equation, we have:

p = (hc)/λ

Now we can calculate the momentum for each option given:

(a) p = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (2.24 x 10^-28 kg) = 8.85 x 10^-6 kg·m/s

(b) p = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (3.94 x 10^-28 kg) = 4.49 x 10^-6 kg·m/s

(c) p = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (5.54 x 10^-28 kg) = 3.33 x 10^-6 kg·m/s

(d) p = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (8.14 x 10^-28 kg) = 2.27 x 10^-6 kg·m/s

(e) p = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / (9.94 x 10^-28 kg) = 1.81 x 10^-6 kg·m/s

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A certain lightbulb is rated at 60.0W when operating at an rams voltage of 120V. (a) What is the peak voltage applied across the bulb?

Answers

The power rating (P) of a certain lightbulb is 60.0W when operating at an rms voltage of 120V.

We are to determine the peak voltage (Vp) applied across the bulb.There is a direct relationship between the root-mean-square (rms) value and peak value of a sinusoidal alternating current (AC) waveform.

Peak value is equal to the square root of 2 times the rms value.Therefore, peak voltage (Vp) can be calculated as follows:Vp = √2 × Vrms Hence, Peak voltage (Vp) applied across the bulb ≈ 1.414 × 120V = 169.7 VAnswer: 169.7 V

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Two identical positive charges exert a re- pulsive force of 6.3 x 10-9 N when separated by a distance 3.9 × 10-10 m. Calculate the charge of each. The Coulomb constant is 8.98755 x 10⁹ Nm²/C². Answer in units of C.

Answers

The charge of each identical positive charge is 9 x 10⁻¹⁰ C.

The electrostatic-force between two charges can be calculated using Coulomb's law:

F = (k * |q₁ * q₂|) / r²

Where:

F is the electrostatic force

k is the Coulomb constant (8.98755 x 10⁹ Nm²/C²)

q₁ and q₂ are the charges of the two charges

r is the distance between the charges

In this case, we are given:

F = 6.3 x 10⁻⁹ N

r = 3.9 x 10⁻¹⁰ m

k = 8.98755 x 10⁹ Nm²/C²

Plugging in the values into Coulomb's law equation:

6.3 x 10⁻⁹ N = (8.98755 x 10⁹ Nm²/C² * |q₁ * q₂|) / (3.9 x 10⁻¹⁰ m)²

Simplifying the equation, we can substitute |q₁ * q₂| with q², as the charges are identical:

6.3 x 10⁻⁹ N = (8.98755 x 10⁹ Nm²/C² * q²) / (3.9 x 10⁻¹⁰ m)²

Solving for q, we find:

q² = (6.3 x 10⁻⁹ N * (3.9 x 10⁻¹⁰ m)²) / (8.98755 x 10⁹ Nm²/C²)

q² = 8.1 x 10⁻¹⁹ C²

Taking the square root of both sides to solve for q, we get:

q = ± 9 x 10⁻¹⁰ C

Since the charges are positive, the charge of each identical positive charge is 9 x 10⁻¹⁰ C.

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The electric field of an electromagnetic wave traveling in vacuum is described by the
following wave function:
E = 5 cos[kx - (6.00 × 10^9)t]j
where k is the wavenumber in rad/m, x is in m, r is in s. Find the following quantities:
a. amplitude
b. frequency
c. wavelength
d. the direction of the travel of the wave
e. the associated magnetic field wave

Answers

The electric field wave has an amplitude of 5, a frequency of 6.00 × 10^9 Hz, a wavelength determined by the wavenumber k, travels in the j direction, and is associated with a magnetic field wave.

The amplitude of the wave is the coefficient of the cosine function, which in this case is  The frequency of the wave is given by the coefficient in front of 't' in the cosine function, which is 6.00 × 10^9 rad/s. Since frequency is measured in cycles per second or Hertz (Hz), the frequency of the wave is 6.00 × 10^9 Hz.

The wavelength of the wave can be determined from the wavenumber (k), which is the spatial frequency of the wave. The wavenumber is related to the wavelength (λ) by the equation λ = 2π/k. In this case, the given wave function does not explicitly provide the value of k, so the specific wavelength cannot be determined without additional information.

The direction of travel of the wave is given by the direction of the unit vector j in the wave function. In this case, the wave travels in the j-direction, which is the y-direction.

According to Maxwell's equations, the associated magnetic field (B) wave can be obtained by taking the cross product of the unit vector j with the electric field unit vector. Since the electric field is given by E = 5 cos[kx - (6.00 × 10^9)t]j, the associated magnetic field is B = (1/c)E x j, where c is the speed of light. By performing the cross-product, the specific expression for the magnetic field wave can be obtained.

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A ray of light origimates in glass and travels to ain. The angle of incidence is 36∘. The ray is partilly reflected from the interfece of gloss and oin at the anple θ2​ and refrocted at enfle θ3​. The index of refraction of the gless is 1.5. a) Find the speed of light in glass b) Find θ2​ c) Find θ3​ d). Find the critcal ancle

Answers

a) The speed of light in glass can be found using the formula v = c/n, where v is the speed of light in the medium (glass), c is the speed of light in vacuum (approximately 3x10^8 m/s), and n is the refractive index of glass (1.5). Therefore, the speed of light in glass is approximately 2x10^8 m/s.

b) To find θ2​, we can use Snell's law, which states that n1*sin(θ1) = n2*sin(θ2), where n1 is the refractive index of the initial medium (glass), n2 is the refractive index of the second medium (air), and θ1 and θ2 are the angles of incidence and reflection, respectively. Given that θ1 is 36∘ and n1 is 1.5, we can solve for θ2:

1.5*sin(36∘) = 1*sin(θ2)

θ2 ≈ 23.49∘

c) To find θ3​, we can use Snell's law again, but this time with the refractive index of air (approximately 1) and the refractive index of glass (1.5). Given that θ2 is 23.49∘ and n1 is 1.5, we can solve for θ3:

1*sin(23.49∘) = 1.5*sin(θ3)

θ3 ≈ 15.18∘

d) The critical angle is the angle of incidence at which the refracted angle becomes 90∘. Using Snell's law with n1 (glass) and n2 (air), we can find the critical angle (θc):

n1*sin(θc) = n2*sin(90∘)

1.5*sin(θc) = 1*sin(90∘)

θc ≈ 41.81∘

Therefore, the critical angle is approximately 41.81∘.

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What is the strength of the electric field between two parallel
conducting plates separated by 1.500E+0 cm and having a potential
difference (voltage) between them of 12500 V?

Answers

The strength of the electric field between the two parallel conducting plates is 8333.33 V/m.

The strength of the electric field between two parallel conducting plates can be calculated using the formula:

E = V / d

Given:

Voltage (V) = 12500 V

Separation distance (d) = 1.500E+0 cm = 1.500 m (converted to meters)

Now we can calculate the electric field strength (E) using the given values:

E = 12500 V / 1.500 m

After calculating the values, the electric field strength between the plates is approximately 8,333.33 V/m.

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The refraction of light is that physical phenomenon by which
light, when passing from one medium to another, deviates from its
original direction.
Select one:
True
False

Answers

The statement "The refraction of light is that physical phenomenon by which light, when passing from one medium to another, deviates from its original direction" is true.

When a beam of light passes from one transparent medium to another, such as from air to water or from water to glass, it bends or deviates from its original path. This bending of light is called refraction. The angle of incidence, the refractive index of the medium, and the angle of refraction determine the amount of bending.

A substance's refractive index, or index of refraction, is a measure of how much the speed of light changes when it travels through it. Light travels faster in a medium with a lower refractive index than in a medium with a higher refractive index.

The amount of bending is determined by the ratio of the speed of light in a vacuum to the speed of light in a medium, known as the refractive index. The refractive index of a substance determines the degree to which light is refracted when it passes through it.

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Prove that in a normed vector space the only sets that are open
and closed at the same time are the empty set and space.

Answers

To prove that in a normed vector space, the only sets that are open and closed at the same time are the empty set and space, we can use the following proof.

X be a normed vector space, and let A be a subset of X that is both open and closed.Let x be an element of A. Since A is open, there exists an open ball centered at x, denoted by B(x, r), that is contained in A. Since A is closed, its complement, X - A, is also open.

There exists an open ball centered at x, denoted by B(x, s), that is contained in X - A. We can choose r and s such that r + s < d(x, X - A), where d denotes the distance function in X.

B(x, r) and B(x, s) are disjoint and contained in A and X - A, respectively.

Consider the sequence {y_n} defined by y_n = x + (r/2^n)v for n = 1, 2, ... , where v is a unit vector in X. Note that the sequence {y_n} is contained in B(x, r), and hence in A.

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An 13.9-kg stone at the end of a steel (Young's modulus 2.0 x 10¹1 N/m²) wire is being whirled in a circle at a constant tangential speed of 11.1 m/s. The stone is moving on the surface of a frictionless horizontal table. The wire is 3.24 m long and has a radius of 1.42 x 10³ m. Find the strain in the wire

Answers

The strain in the wire is 3.1 x 10⁻⁴ or 0.00031 or 0.031%. This means that the steel wire is stretched by 0.031% due to the weight of the stone and the circular motion.

Mass of the stone, m = 13.9 kg

Speed of the stone, v = 11.1 m/s

Length of the wire, L = 3.24 m

Radius of the wire, r = 1.42 x 10³ m

Young's modulus of steel wire, Y = 2.0 x 10¹¹ N/m²

Formula used:

Strain, ε = (FL)/AY

where, F is the force applied

L is the length of the wire

A is the area of cross-section of the wire

Y is the Young's modulus of the wire

For a wire moving in a horizontal circle, the tension, T in the wire is given by

T = mv²/r

where, m is the mass of the stone

v is the speed of the stoner is the radius of the circle

Substituting the given values, we get:

T = (13.9 kg) x (11.1 m/s)² / (1.42 x 10³ m)

   = 15.9 NA

s the stone is moving on a frictionless surface, the only force acting on the stone is the tension in the wire. Hence, the tension in the wire is also equal to the force acting on it. Therefore, we use T in place of F to calculate the strain.

ε = (T x L) / (A x Y)

We need to find ε.

Solving for ε, we get:

ε = (T x L) / (A x Y)

  = (15.9 N x 3.24 m) / [(π x (1.42 x 10⁻³ m)²)/4 x (2.0 x 10¹¹ N/m²)]

  = 3.1 x 10⁻⁴ or 0.00031 or 0.031%

Therefore, the strain in the wire is 3.1 x 10⁻⁴ or 0.00031 or 0.031%. This means that the steel wire is stretched by 0.031% due to the weight of the stone and the circular motion.

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A model airplane with mass 0.750 kg is tethered to the ground by a wire so that it flies in a horizontal circle 30.0m in radius. The airplane engine provides a net thrust of 0.800N perpendicular to the tethering wire.(b) Find the angular acceleration of the airplane.

Answers

The angular acceleration of the airplane is 0.0356 rad/s².

To find the angular acceleration of the airplane, we can use the equation:

Net force = mass × radius × angular acceleration

Given that the net force is 0.800N and the mass of the airplane is 0.750 kg, we can rearrange the equation to solve for angular acceleration.

Angular acceleration = Net force / (mass × radius)

Substituting the given values:

Angular acceleration = 0.800N / (0.750 kg × 30.0m)

Calculating this gives us:

Angular acceleration = 0.800N / 22.5 kg·m/s²

Simplifying further, the angular acceleration is:

Angular acceleration = 0.0356 rad/s²

Therefore, the angular acceleration of the airplane is 0.0356 rad/s². This means that the airplane is accelerating angularly at a rate of 0.0356 radians per second squared..

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Consider the nuclear fusion reaction 12​H+12​H−>13​H+11​H Each fusion event releases approximately 4.03MeV of energy. How much total energy, in joules, would be released if all the deuterium atoms (12​H) in a typical 0.290 kg glass of water were to undergo this fusion reaction? Assume that approximately 0.0135% of all the hydrogen atoms in the water are deuterium. energy released: Incorrect A typical human body metabolizes energy from food at a rate of about 104.5 W, on average. How long, in days, would it take a human to metabolize the amount of energy released? time to metabolize the amount of energy released: days

Answers

To calculate the total energy released in the fusion reaction and the time it would take for a human to metabolize that energy, we need to determine the number of deuterium atoms in the given mass of water and then use the conversion factors to calculate the energy and time.

Given:

Mass of water (m) = 0.290 kg

Energy released per fusion event (E) = 4.03 MeV

Percentage of deuterium atoms in water = 0.0135%

Average human metabolic rate (P) = 104.5 W

Calculate the number of deuterium atoms in the mass of water:

Number of deuterium atoms (N) = (0.0135/100) * (6.022 × 10^23) * (0.290 kg / (2.014 g/mol))

N ≈ 1.051 × 10^19 atoms

Calculate the total energy released:

Total energy released (E_total) = N * E * (1.602 × 10^-13 J/MeV)

E_total ≈ 1.051 × 10^19 * 4.03 * (1.602 × 10^-13) J

E_total ≈ 6.78 × 10^5 J

Calculate the time to metabolize the energy:

Time (t) = E_total / P

t ≈ 6.78 × 10^5 J / 104.5 W

t ≈ 6492 s

Convert seconds to days:

t ≈ 6492 s / (24 * 60 * 60 s/day)

t ≈ 0.0752 days

The total energy released if all the deuterium atoms in a typical 0.290 kg glass of water undergo fusion is approximately 6.78 × 10^5 J.

It would take approximately 0.0752 days for a human to metabolize that amount of energy.

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An examination reveals that a patient cannot clearly see any object that lies closer than 58.0 cm to the patient's eye. (a) Which of the following terms best describes this distance? magnification focal length far point near point an ideal thin lens, which lies adjacent to the eye. cm (c) What is the power, P, of the contact lens (in diopters)? x diopters

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An examination reveals that a patient cannot clearly see any object that lies closer than 58.0 cm to the patient's eye. (a) Which of the following terms best describes this distance  Magnification, focal length, far point, near point, and an ideal thin lens, which lies adjacent to the eye. The near point term best describes this distance.

The near point is the minimum distance from the eye at which the eye can see objects clearly. The near point, which is sometimes referred to as the closest point of clear vision, is the shortest distance from the eye that a person can focus on an object with their natural lens. As one gets older, the ability of the eye to focus at near distances declines.

(c) What is the power, P, of the contact lens (in diopters)? Let's first calculate the focal length, which is f = 1/do - 1/di We know that do = infinity (since the object is at infinity).

So,f = 1/di  => di = 1/f = 0.0345 m  Using the lens formula, we get:  1/f = 1/di - 1/do We know that di = -58 cm = -0.58 m and do = infinity.So,1/f = -1/0.58  => f = -1.72 m The power of the lens, P = 1/f = -0.58 diopters (negative sign indicates that the lens is a concave lens). Therefore, the power of the contact lens is -0.58 diopters.

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Question 43 1 pts In what form does water exist on the Moon? There is water ice in the bright regions of the lunar maria. There are shallow lakes of liquid water in the deepest craters. There are small pools of liquid water just beneath the surface. There is no water in any form on the Moon There is water ice in craters near the poles.

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Water exists on the Moon in the form of water ice in craters near the poles.

Scientific studies and observations have provided evidence for the presence of water ice on the Moon. The lunar poles, specifically the permanently shadowed regions within craters, are known to harbor water ice.

These regions are characterized by extremely low temperatures and lack of sunlight, allowing ice to persist. The ice is believed to have originated from various sources, including cometary impacts and the solar wind, which carried hydrogen that could react with oxygen to form water molecules.

NASA's Lunar Reconnaissance Orbiter (LRO) mission and other spacecraft have provided valuable data on the presence of water ice. LRO's instruments, such as the Lunar Exploration Neutron Detector (LEND), have detected elevated levels of hydrogen at the poles, indicating the presence of water ice.

Additionally, the Lunar Crater Observation and Sensing Satellite (LCROSS) mission performed an impact experiment, confirming the presence of water ice in a permanently shadowed crater.

The discovery of water ice on the Moon has significant implications for future lunar exploration and potential resource utilization. It provides a potential source of water for sustaining human presence, producing rocket propellant, and supporting other activities.

However, it's important to note that while water ice exists in craters near the poles, it is not distributed across the entire lunar surface, and other regions of the Moon do not possess significant amounts of water in any form.

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1. What makes a spaceship orbit the earth?
a. The velocity makes a spaceship orbit the earth.
b. The gravitational force makes the spaceship to travel in a circular orbit.
c. The thrust makes a spaceship rotates around the earth.
d. Spaceship cannot orbit the earth because of the gravity.
2. What is the difference between evaporating and boiling?
a. Boiling is not evaporating because the temperature of boiling is higher than that of evaporating.
b. Evaporating happens only on the top surface of liquid while boiling happens both on top surface of liquid and within the liquid.
c. Boiling is one kind of evaporating, so they are the same for water.
d. Evaporating is fast than boiling.
3. Why do some clothes cling while others repel?
a. Like charges attract and opposite charges repel.
b. Like charges repel and opposite charges attract.
c. Charges attach at larger distance and reply when they are close.
d. none of the above

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The gravitational force (b) is what allows a spaceship to orbit the Earth, keeping it in a circular path.

Evaporating (b) occurs only on the liquid's surface, while boiling happens both on the surface and within the liquid.

Clothes cling or repel based on material properties, not electric charges (d). It's not related to electrical attraction or repulsion.

1.  (b) The gravitational force makes the spaceship travel in a circular orbit. In orbit, the gravitational force between the spaceship and the Earth keeps the spaceship moving in a curved path around the Earth, creating a stable orbit.

2.(b) Evaporating happens only on the top surface of a liquid, while boiling occurs both on the top surface and within the liquid.

Evaporation is a process in which molecules at the liquid's surface gain enough energy to escape into the surrounding space, while boiling involves the rapid vaporization of a liquid throughout the entire volume due to the input of heat.

3.(d) None of the above. The cling or repel of clothes is not related to electric charges. It is primarily determined by the materials and their surface properties, such as their ability to generate static electricity or their surface tension.

The main factors for a spaceship to orbit the Earth are the gravitational force, while the difference between evaporating and boiling lies in the extent of the process within the liquid. The cling or repel of clothes is determined by material properties rather than electrical charges.

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"A coil with 450 turns is exposed to a magnetic flux (see picture). The flow through the coil cross section increases by 1.5 miliweber per second.
a) Determine the voltage induced in the coil.

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The number of turns in a coil is 450, and the magnetic flux passing through the coil cross-section increases at a rate of 1.5 mWb/s, we need to determine the voltage induced in the coil using Faraday's law of electromagnetic induction.

What is Faraday's law of electromagnetic induction? Faraday's law of electromagnetic induction states that the rate of change of magnetic flux through a closed loop induces an electromotive force (emf) and a corresponding electrical current in the loop. The induced electromotive force is directly proportional to the rate of change of magnetic flux through the loop.

Mathematically, Faraday's law of electromagnetic induction can be expressed as; EMF = -dΦ/dt where, EMF is the electromotive force (V),dΦ is the change in magnetic flux through the coil cross-section (Wb), and dt is the change in time (s).Therefore, the voltage induced in the coil is given by; EMF = -dΦ/dtEMF = -1.5 mWb/s * 450EMF = -675 V. Thus, the voltage induced in the coil is -675 V. The negative sign indicates that the voltage is induced in the opposite direction to the change in magnetic flux.

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A 4 foot, 2.65 inch by 1 foot, 10.96 inch steel panel is heated from 9 C to 60 C. Calculate the change in surface area due to the temperature change. Report your answer in square inches rounded to 2 decimal places with units.

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The change in surface area due to temperature change is 0.71 in² (approx).

Let's calculate the change in surface area due to temperature change. We can use the formula below:

ΔA = αA_0 ΔT

where, ΔA = change in surface area due to temperature changeα = coefficient of thermal expansion

A_0 = initial surface area

ΔT = change in temperature

Substitute the given values, ΔT = 60°C - 9°C = 51°C= 4 × 12 + 2.65 = 50.65 inches (length)= 1 × 12 + 10.96 = 22.96 inches (breadth)

A_0 = length × breadth= 50.65 × 22.96 = 1164.86 in²

Coefficient of thermal expansion (α) for steel = 1.2 × 10⁻⁵/°CΔA = αA_0 ΔT= (1.2 × 10⁻⁵/°C)(1164.86 in²)(51°C)= 0.71404 in² (approx)

Therefore, the change in surface area due to temperature change is 0.71 in² (approx).

We are given a steel panel of dimensions 4 feet, 2.65 inches by 1 foot, and 10.96 inches. It is heated from 9 C to 60 C and we are required to find the change in surface area due to temperature change. We have to calculate the change in surface area due to the expansion of the steel panel caused by the increase in temperature. This is given by the formula ΔA = αA_0 ΔT, where ΔA is the change in surface area, α is the coefficient of thermal expansion, A_0 is the initial surface area and ΔT is the change in temperature. We first convert the given dimensions from feet and inches to inches only. The length is 4 feet × 12 inches per foot + 2.65 inches = 50.65 inches. The breadth is 1 foot × 12 inches per foot + 10.96 inches = 22.96 inches. Using these dimensions, we calculate the initial surface area A_0 as length × breadth which is 1164.86 in². The coefficient of thermal expansion for steel is 1.2 × 10⁻⁵/°C. The change in temperature ΔT is calculated as 60°C - 9°C = 51°C. Substituting these values in the formula, we get ΔA = (1.2 × 10⁻⁵/°C)(1164.86 in²)(51°C) = 0.71404 in². Therefore, the change in surface area due to temperature change is 0.71 in² (approx).

Therefore, the change in surface area due to temperature change is 0.71 in² (approx).

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You illuminate a slit with a width of 75.1 μm75.1 μm with a light of wavelength 727 nm727 nm and observe the resulting diffraction pattern on a screen that is situated 2.23 m2.23 m from the slit. What is the width, in centimeters, of the pattern's central maximum?

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The width of the central maximum of the diffraction pattern is approximately 4.82 cm.

The width of the central maximum of a diffraction pattern can be determined using the formula:

w = (λ * D) / d

where:

w is the width of the central maximum,

λ is the wavelength of light,

D is the distance between the slit and the screen, and

d is the width of the slit.

In this case, the width of the slit is given as 75.1 μm (or 75.1 × 10^(-6) m) and the wavelength of light is 727 nm (or 727 × 10^(-9) m). The distance between the slit and the screen is 2.23 m.

Substituting these values into the formula:

w = (727 × 10^(-9) m * 2.23 m) / (75.1 × 10^(-6) m)

Simplifying the expression:

w = (1.62 × 10^(-6) m * 2.23 m) / (75.1 × 10^(-6) m)

≈ 0.0482 m

Converting the width to centimeters:

w ≈ 0.0482 m * 100 cm/m

≈ 4.82 cm

Therefore, the width of the central maximum of the diffraction pattern is approximately 4.82 centimeters.

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As an object falls freely downward with negligible air resistance, its (b) acceleration increases (a) velocity increases neither a nor b both \( a \) and \( b \)

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When an object falls freely downward with negligible air resistance, its acceleration increases.

The acceleration of a freely falling object near the surface of the Earth is due to the force of gravity acting on it. According to Newton's second law of motion, the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = m * a). In this case, the only significant force acting on the object is the force of gravity, given by the equation F = m * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2 near the surface of the Earth).

As an object falls freely downward, the force of gravity remains constant, as the mass of the object does not change. Therefore, the net force acting on the object is constant. According to Newton's second law, since the net force is constant and the mass of the object remains the same, the acceleration of the object must also be constant.

In conclusion, when an object falls freely downward with negligible air resistance, its acceleration remains constant throughout the fall. Thus, the correct answer is "neither a nor b."

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A battery with an emf of 60 V is connected to the two Part A capacitors shown in the figure(Figure 1). Afterward, the charge on capacitor 2 is 270μC. What is the capacitance of capacitor 2 ? Express your answer using two significant figures. Figure 1 of 1 X Incorrect; Try Again; 4 attempts remaining

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The capacitance of capacitor 2 is approximately X μF (two significant figures).

To find the capacitance of capacitor 2, we can use the formula for the charge on a capacitor: Q = CV, where Q is the charge, C is the capacitance, and V is the voltage (emf) across the capacitor.

Given that the emf of the battery is 60 V and the charge on capacitor 2 is 270 μC, we can rearrange the formula as follows:

270 μC = C × 60 V

To find the capacitance C, we divide both sides of the equation by 60 V:

C = (270 μC) / (60 V)

Simplifying, we get:

C ≈ 4.5 μF

Therefore, the capacitance of capacitor 2 is approximately 4.5 μF, rounded to two significant figures.

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Score 1 Starting from rest, a turnable rotates at angular acceleration of 0.13 rad/s2. How long does it take for it speed to get to 6 rad/s? 3A 1110 kg car traveling clockwise at a constant speed along a flat horizontal circular track of radius 26 m. The car takes 21 s to complete one lap around the track. What is the magnitude of the force of friction exerted on the car by the track? The angular velocity of a rotating object is defined by the function w = 4t³ - 2t + 3 What is the objects angular acceleration at t = 5 seconds?

Answers

The angular acceleration at t = 5 seconds is 298 rad/s².

Angular acceleration, α = 0.13 rad/s²

Initial angular velocity,

ω₁ = 0Final angular velocity,

ω₂ = 6

We have to find the time it takes to reach this final velocity. We know that

Acceleration, a = αTime, t = ?

Initial velocity, u = ω₁Final velocity, v = ω₂Using the formula v = u + at

The final velocity of an object, v = u + at is given, where v is the final velocity of the object, u is the initial velocity of the object, a is the acceleration of the object, and t is the time taken for the object to change its velocity from u to v.

Substituting the given values we get,

6 = 0 + (0.13)t6/0.13 = t461.5 seconds ≈ 62 seconds

Therefore, the time taken to get to 6 rad/s is 62 seconds.3) The given parameters are given below:

Mass of the car, m = 1110 kg

Radius of the track, r = 26 m

Time taken to complete one lap around the track, t = 21 sWe have to find the magnitude of the force of friction exerted on the car by the track.

We know that:

Centripetal force, F = (mv²)/r

The force that acts towards the center of the circle is known as centripetal force.

Substituting the given values we get,

F = (1110 × 6.12²)/26F

= 16548.9 N

≈ 16550 N

To find the force of friction, we have to find the force acting in the opposite direction to the centripetal force.

Therefore, the magnitude of the force of friction exerted on the car by the track is 16550 N.2) The given angular velocity function is, ω = 4t³ - 2t + 3We have to find the angular acceleration at t = 5 seconds.We know that the derivative of velocity with respect to time is acceleration.

Therefore, Angular velocity, ω = 4t³ - 2t + 3 Angular acceleration, α = dω/dt Differentiating the given function w.r.t. t we get,α = dω/dt = d/dt (4t³ - 2t + 3)α = 12t² - 2At t = 5,α = 12(5²) - 2 = 298 rad/s².

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A piano wire of linear mass density 0.0050 kg/m is under a tension of 1350 N. What is the wave speed in this wire? O 1040 m/s O 260 m/s O 520 m/s 130 m/s Moving to another question will save this resp

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The wave speed in the piano wire, under a tension of 1350 N and linear mass density of 0.0050 kg/m, is approximately 520 m/s.

To calculate the wave speed in the piano wire, we can use the formula:

Wave speed (v) = sqrt(Tension (T) / linear mass density (μ))

Given:

Linear mass density (μ) = 0.0050 kg/m

Tension (T) = 1350 N

Substituting these values into the formula, we get:

Wave speed (v) = sqrt(1350 N / 0.0050 kg/m)

Wave speed (v) = sqrt(270,000 m²/s² / kg/m)

Wave speed (v) = sqrt(270,000) m/s

Wave speed (v) ≈ 519.62 m/s

Therefore, the wave speed in the piano wire is approximately 520 m/s.

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A balloon charged with static electricity will stick to an insulating wall because
a.) The charges in the balloon polarize the charges in the wall
b.) None of these, the balloon will not stick to an insulating surface
c.) The strong nuclear force holds the balloon when the atomic nuclei get close
d.) Gravity pulls the atoms in the balloon towards the atoms in the wall

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option a) is the correct answer.

a) The charges in the balloon polarize the charges in the wall.

When a balloon is charged with static electricity, it gains either an excess of positive or negative charges. These charges create an electric field around the balloon. When the charged balloon is brought close to an insulating wall, such as a wall made of plastic or glass, the charges in the balloon polarize the charges in the wall.

The positive charges in the balloon attract the negative charges in the wall, and the negative charges in the balloon attract the positive charges in the wall. This polarization creates an attractive force between the balloon and the wall, causing the balloon to stick to the insulating surface.

Therefore, option a) is the correct answer.

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how much does a 1 kg pineapple weigh on earth.

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A 1 kg pineapple weighs approximately 9.8 Newtons on Earth. The weight of an object is determined by the force of gravity acting on it, and on Earth, the acceleration due to gravity is approximately 9.8 m/s^2.

The weight of an object is the force exerted on it due to gravity. It is measured in Newtons (N) and is directly proportional to the mass of the object. On Earth, the acceleration due to gravity is approximately 9.8 m/s^2.

This means that for every kilogram of mass, an object experiences a gravitational force of 9.8 Newtons.

In the case of a 1 kg pineapple on Earth, its weight can be calculated by multiplying its mass (1 kg) by the acceleration due to gravity (9.8 m/s^2):

Weight = Mass × Acceleration due to gravity

Weight = 1 kg × 9.8 m/s^2

Therefore, a 1 kg pineapple weighs approximately 9.8 Newtons on Earth.

It's important to note that weight can vary depending on the gravitational force of the celestial body. For example, on the Moon, where the acceleration due to gravity is much lower than on Earth, the same 1 kg pineapple would weigh less.

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The output period of a frequency division circuit that contains 4 flip-flops with an input clock frequency of 80 MHz is: a) 25 ns b) 50 ns c) 125 ns d) 200 ns e) None

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The output period of a frequency division circuit that contains 4 flip-flops with an input clock frequency of 80 MHz is 200 ns. The correct option is D.

A frequency division circuit is an electronic circuit that divides the input signal frequency by an integer factor and produces an output signal. Flip-flops are used in frequency dividers to provide clock signals to the succeeding flip-flop.

What is frequency division?

Frequency division is a process of converting an input signal of one frequency to an output signal of a different frequency that is a submultiple of the input signal frequency. The frequency division ratio is equal to the number of input signal cycles required to produce one output cycle.

Input clock frequency = 80 MHz

Number of flip-flops = 4

The output frequency of the circuit is equal to the input frequency divided by the frequency division ratio (FDR), which is equal to 2 to the power of the number of flip-flops.

Expressed in mathematical terms,

FDR = 2⁴ = 16

Output frequency = Input frequency / FDR= 80 MHz / 16 = 5 MHz

Output period = 1 / output frequency= 1 / 5 MHz= 200 ns

Therefore, the correct option is D, which is 200 ns.

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The inductance of a closely packed coil of 420 turns is 11 mH.
Calculate the magnetic flux through the coil when the current is
4.7 mA.

Answers

The magnetic flux through the coil when the current is 4.7 mA is approximately 21.714 milliWebers (mWb).

The magnetic flux through a coil can be calculated using the formula:

Φ = L * I

where Φ is the magnetic flux, L is the inductance of the coil, and I is the current passing through the coil.

Given:

Number of turns in the coil (N) = 420

Inductance of the coil (L) = 11 mH = 11 × 10^(-3) H

Current passing through the coil (I) = 4.7 mA = 4.7 × 10^(-3) A

First, we need to calculate the effective number of turns by multiplying the number of turns with the current:

[tex]N_eff = N * IN_eff = 420 * 4.7 × 10^(-3)N_eff = 1.974\\[/tex]

Now, we can calculate the magnetic flux using the formula:

[tex]Φ = L * IΦ = (11 × 10^(-3) H) * (1.974)Φ = 21.714 × 10^(-3) WbΦ = 21.714 mWb\\[/tex]

Therefore, the magnetic flux through the coil when the current is 4.7 mA is approximately 21.714 milliWebers (mWb).

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