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3 years ago

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Create two models to compare the amounts of potential and kinetic energy in the Tuesday and Wednesday spacecraft launches. Use a

magnet to represent the spacecraft in each launch. Use the symbols shown in the Key in your models.

1 answer:

Lapatulllka [165]3 years ago

6 0

Answer:

DUDE USE UR BRAOIIINNNNNNNNNNNNN ITS B

Explanation:

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If I measure the current to be 0.5 A and the voltage is 4.5 V. What is my Power?

Korvikt [17]

As we know that P=IV so P = (0.5A)(4.5V)
P= 2.25 watt

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3 years ago

You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with

Misha Larkins [42]

Answer:

Sam's and Abigail speeds before colliding were a. 12.34 m/s and b. 2.86 m/s, respectively. Their total kinetic energy was diminished by c. 1484.42 J, approximately

Explanation:

By conservation of momentum, we have

$\Delta \vec{p} = 0\\\\m\vec{v}_i = m\vec{v}_f\\m_S v_S\hat{i} + m_A v_A\hat{j} = m_S \times 7 \cos{(40)} \hat{i} + m_S \times 7 \sin{(40)} \hat{j} + m_A \times 9.4 \cos{(18)} \hat{i} - m_A \times 9.4 \sin{(18)} \hat{j}.$

Writing for each direction at a time,

$m_S v_S = 7m_S \cos{(40)} +9.4 m_A \cos{(18)}\\v_S = 7 \cos{(40)} + 9.4 \frac{m_A}{m_S} \cos{(18)} = 7 \cos{(40)} + 9.4\times\frac{57}{73} \cos{(18)} \approx \mathbf{12.34}\\m_A v_A = 7m_S\sin{(40)} - 9.4m_A\sin{(18)}\\v_A = 7\frac{m_S}{m_A}\sin{(40)} - 9.4\sin{(18)} = 7\times\frac{73}{57}\sin{(40)} - 9.4\sin{(18)} \approx \mathbf{2.86}.$

Their kinetic energy changed by

$K_f - K_i = \left( \frac{1}{2}m_s v_{fs}^2 + \frac{1}{2}m_a v_{fa}^2 \right) - \left( \frac{1}{2}m_s v_{is}^2 + \frac{1}{2}m_a v_{ia}^2 \right) = \left( \frac{1}{2}\times 73 \times 7^2 + \frac{1}{2}\times 57 \times 9.4^2 \right) - \left( \frac{1}{2}\times 73 \times 12.34^2 + \frac{1}{2}\times 57 \times 2.86^2 \right) \approx \mathbf{-1484.418 J}.$

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3 years ago

Assume that a person bouncing a ball represents a closed system. Which statement best describes how the amounts of the ball's po

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Answer:

C.

Explanation:

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What is the name of the force that keeps the planets and other celestial bodies orbiting around the sun

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Gravity is the name of the force that keeps the planets in orbit

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3 years ago

Read 2 more answers

The 1.19 kg head of a hammer has a speed of 8.4 m/s just before it strikes a nail (Fig. 14-13) and is brought to rest. Estimate

KATRIN_1 [288]

Answer:

The temperature rise is $143.9\ K$ .

Explanation:

Given the mass of the hammer $(m)$ is 1.19 kg.

And the speed of the hammer $(v)$ is 8.4 m/s.

We need to find the temperature change of iron nail having 20 hammer blows.

First, we will find the kinetic energy of the each blow of hammer.

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}\times 1.19\times 8.4^2=41.99\ J$

For 20 such blows, the kinetic energy will be

$20\times K.E=20\times 41.99=839.8\ J$

Let us assume that the nail absorbs all the kinetic energy. So, this 839.8 J will be converted into heat energy $(Q)$ .

Now,

$Q=839.8\ J$

Also, we know that

$Q=mc\Delta_T$

Where,

$m$ is the mass of the nail, which is $13\ g=\frac{13}{1000}=0.013\ kg$

$c$ is the specific heat capacity of iron, which is 449 J/kg.K

$\Delta_T$ is the change in temperature.

Plug these value we get,

$Q=mc\Delta_T\\839.8\ J=(0.013\ kg)(449\ J/kg.K)\times\Delta_T\\\Delta_T=\frac{839.8}{0.013\times 449}\\\\\Delta_T=143.9\ K$

We can see the temperature rise is 143.9 K.

4 0

4 years ago