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Fudgin [204]
3 years ago
12

Puck B has twice the mass of puck A. Starting from rest, both pucks are pulled the same distance across frictionless ice by stri

ngs with the same tension.a. Compare the final kinetic energies of pucks A and B. b. Compare the final speeds of pucks A and B.
Physics
1 answer:
gogolik [260]3 years ago
8 0

Answer:

(a) 1 : 2

(b) same

Explanation:

Let the mass of puck A is m and the mass of puck B is 2 m.

initial speed for both the pucks is same as u and the distance is same for both is s.

let the tension is T for same.

The kinetic energy is given by

K = 0.5 mv^2

(a) As the speed is same, so the kinetic energy depends on the mass.

So, kinetic energy of A : Kinetic energy of B = m : 2m  = 1 : 2

(b) A the distance s same so the final velocities are also same.

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Answer:

<em>v = 381 m/s</em>

Explanation:

<u>Linear Speed</u>

The linear speed of the bullet is calculated by the formula:

\displaystyle v=\frac{x}{t}

Where:

x = Distance traveled

t = Time needed to travel x

We are given the distance the bullet travels x=61 cm = 0.61 m. We need to determine the time the bullet took to make the holes between the two disks.

The formula for the angular speed of a rotating object is:

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Where θ is the angular displacement and t is the time. Solving for t:

\displaystyle t=\frac{\theta}{\omega}

The angular displacement is θ=14°. Converting to radians:

\theta=14*\pi/180=0.2443\ rad

The angular speed is w=1436 rev/min. Converting to rad/s:

\omega = 1436*2\pi/60=150.3776\ rad/s

Thus the time is:

\displaystyle t=\frac{0.2443\ rad}{150.3776\ rad/s}

t = 0.0016 s

Thus the speed of the bullet is:

\displaystyle v=\frac{0.61}{0.0016}

v = 381 m/s

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A person pushes a 125 kg object initially at rest on a frictionless surface. The object ends up with a final velocity of 1.40 m/
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Answer:

a person pushes a 125 kg

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Waves diffract the most when their wavelength is
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<h2>Answer: about the same size of the gap  or slit</h2>

Diffraction happens when a wave (mechanical or electromagnetic wave, in fact, any wave) meets an obstacle or a slit .When this occurs, the wave bends around the corners of the obstacle or passes through the opening of the slit that acts as an obstacle, forming multiple patterns with the shape of the aperture of the slit.

Note that the principal condition for the occurrence of this phenomena is that the obstacle must be comparable in size (similar size) to the size of the wavelength.

In other words, when the gap (or slit) size is larger than the wavelength, the wave passes through the gap and does not spread out much on the other side, but when the gap size is equal to the wavelength, maximum diffraction occurs.

Therefore:

<h2>Waves diffract the most when their wavelength is <u>about the same size of the gap </u></h2>

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