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Usimov [2.4K]
2 years ago
11

A hockey puck with a mass of 0.17 kg is traveling to the right along the ice at 15 m/s. it strikes a second hockey puck with a m

ass 0.11 kg. the first hockey puck comes to rest after the collision. What is the velocity of the second hockey puck after the collision? ( Round your answer to the nearest integer.)
Physics
1 answer:
Igoryamba2 years ago
3 0

Answer:

v_2 = 23.182m/s

Explanation:

Given

m_1 = 0.17kg

m_2 = 0.11kg

u_1 = 15m/s -- Initial Velocity of the first

u_2 = 0m/s -- Initial Velocity of the second

v_1 = 0m/s -- Final Velocity of the first

Required

Determine the final velocity of the second hockey puck (v_2)

This question illustrates law of conservation of momentum and it'll be solved using the following formula:

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 ---- law of conservation of momentum

Substitute in the right values

0.17 * 15 + 0.11*0 = 0.17 * 0 + 0.11*v_2

0.17 * 15 + 0 = 0 + 0.11*v_2

2.55 = 0.11*v_2

Solve for v2

v_2 = 2.55/0.11

v_2 = 23.182m/s

<em>Hence, the final velocity of the second hockey puck is 23.182m/s</em>

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Leviafan [203]

Answer:

32 C > 32 F > 32 K

Explanation:

32 F, 32 C, 32 K

Let T1 = 32 F

T2 = 32 C

T3 = 32 K

Convert all the temperatures in degree C

The relation between F and C is given by

(F - 32) / 9 = C / 100

so, (32 - 32) / 9 = C / 100

C = 0

So, T1 = 32 F = 0 C

The relation between c and K is given by

C = K - 273 = 32 - 273 = - 241

So, T3 = 32 K = - 241 C

So, T 1 = 0 C, T2 = 32 c, T3 = - 241 C

Thus, T2 > T1 > T3

32C > 32 F > 32 K

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3 years ago
The flow of electricity can be compared of water in
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The flow of electricity can be compared of water in the pipes because both water and electricity moves in the channel.

<h3>How we compare the flow of electricity to water?</h3>

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So we can conclude that the flow of electricity can be compared of water in the pipes because both water and electricity moves in the channel.

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7 0
1 year ago
1. A small light bulb is shining light on a basketball (diameter is 23 cm or 9 inches). The light bulb is 3 m from the closest s
siniylev [52]

Answer:

The size (diameter) of the basketball's shadow on the wall is approximately 53.38 cm

Explanation:

The given parameters of the basketball are;

The diameter of the basketball = 23 cm (9 inches)

The distance of the light bulb from the closest side of the basketball = 3 m

The distance from the ball to the wall = 4 m

The distance from the light source to the center of the ball, d = 3 m + 0.23/2 m = 3.115 m

The angle the light ray makes with the edge of the ball, θ = arctan(0.115/3.115)

Therefore, the ratio of the shadow width divided by 2 to the distance from the light from the wall = 0.115/3.115

The distance from the light from the wall = 3 m + 4 m + 0.23 m = 7.23 m

Therefore;

((The width of the shadow)/2)/(The distance from the light from the wall) = 0.115/3.115

∴ ((The width of the shadow)/2)/(7.23 m) = 0.115/3.115

((The width of the shadow)/2) = 7.23 m × 0.115/3.115 = 16629/62300 m ≈ 0.2669 m = 26.69 cm

The width (diameter) of the shadow on the wall = 2 × 16629/62300 m ≈ 0.5338 m = 53.38 cm

The size (diameter) of the basketball's shadow on the wall ≈ 53.38 cm

4 0
2 years ago
A cart with mass m vibrating at the end of a spring has an extra block added to it when its displacement is x=+A. What should th
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Answer:

The block's mass should be 3m

Explanation:

Given:

Cart with mass m

From the conservation of energy before mass is added,

  \frac{1}{2} mv^{2} = \frac{1}{2} kA^{2}

Where A = amplitude of spring mass system, k = spring constant

  A = v\sqrt{\frac{m}{k} }

Now new mass M is added to the system,

   \frac{1}{2} (m +M ) v^{2}  = \frac{1}{2}  k A^{2}

  A = v \sqrt{\frac{m +M }{k} }

Here, given in question frequency is reduced to half so we can write,

   f' = \frac{f}{2}

Where f = frequency of system before mass is added, f' = frequency of system after mass is added.

        \omega ' = \frac{\omega}{2}

\sqrt{\frac{k}{m +M} }  = \frac{\sqrt{\frac{k}{m} } }{2}

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8 0
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DanielleElmas [232]

Answer:

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Putting all together:

\alpha =\frac{\frac{2\pi}{T_f}-\frac{2\pi}{T_i}}{\Delta t}=\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})

Taking our initial (i) point now and our final (f) point one year later, we would have:

\Delta t=1\ year=(365)(24)(60)(60)s=31536000&#10;s

T_i=0.0786s

T_f=0.0786s+7.03\times10^{-6}s

So for our values we have:

\alpha =\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})=\frac{2\pi}{31536000s}(\frac{1}{0.0786s+7.03\times10^{-6}s}-\frac{1}{0.0786s})=-2.2669642\times^{-10}rad/s^2

Where the minus sign indicates it is decelerating.

8 0
3 years ago
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