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2 years ago

10

The specific heat of copper is 0.39 j/g*degrees celcius how much heat is needed to raise the temperature of 1000.0 g of copper f

rom 25.0 degrees celsius to 45.0 degrees celcius

1 answer:

lapo4ka [179]2 years ago

6 0

In order to calculate the amount of energy needed, we will apply:
heat energy = mass x specific heat capacity x change in temperature
Q = mcΔT

Substituting the given values,
Q = 1000 x 0.39 x (45 - 25)
Q = 7,800 Joules of heat energy are needed

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A block is resting on a platform that is rotating at an angular speed of 2.4 rad/s. The coefficient of static friction between t

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Answer:

r = 0m is the Minimum distance from the axis at which the block can remain in place wothout skidding.

Explanation:

From a sum of forces:

$Ff = m*a$ where Ff = μ * N and $a = \frac{V^2}{r}=\omega^2*r$

N - m*g = 0 So, N = m*g. Replacing everything on the original equation:

$\mu*m*g = m*\omega^2*r$ (eq2)

Solving for r:

$r = \frac{\mu*g}{\omega^2}=1.41m$

If we analyze eq2 you can conclude that as r grows, the friction has to grow (assuming that ω is constant), so the smallest distance would be 0 and the greatest 1.41m. Beyond that distance, μ has to be greater than 0.83.

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2 years ago

The place theory of hearing is better than frequency theory at explaining pitch perception for __________ sounds

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The answer is : Low frequency sounds

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Define Impulse And Momentum<br><br>Plsss Answer Now I Need Now

andrew-mc [135]

Answer:

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1 year ago

In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviationMVC. Suppose a 1

lara31 [8.8K]

Answer:

Part a)

$f = \frac{8}{9}$

Part b)

$f = \frac{120}{169}$

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

Explanation:

Part a)

Let say the collision between Moose and the car is elastic collision

So here we can use momentum conservation

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1000 v_o = 1000 v_{1f} + 500 v_{2f}$

also by elastic collision condition we know that

$v_{2f} - v_{1f} = v_o$

now we have

$2v_o = 2v_{1f} + v_o + v_{1f}$

now we have

$v_{1f} = \frac{v_o}{3}$

Now loss in kinetic energy of the car is given as

$\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)$

$\Delta K = \frac{1}{2}m(v_o^2 - \frac{v_o^2}{9})$

so fractional loss in energy is given as

$f = \frac{\Delta K}{K}$

$f = \frac{\frac{4}{9}mv_o^2}{\frac{1}{2}mv_o^2}$

$f = \frac{8}{9}$

Part b)

Let say the collision between Camel and the car is elastic collision

So here we can use momentum conservation

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1000 v_o = 1000 v_{1f} + 300 v_{2f}$

also by elastic collision condition we know that

$v_{2f} - v_{1f} = v_o$

now we have

$10v_o = 10v_{1f} + 3(v_o + v_{1f})$

now we have

$v_{1f} = \frac{7v_o}{13}$

Now loss in kinetic energy of the car is given as

$\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)$

$\Delta K = \frac{1}{2}m(v_o^2 - \frac{49v_o^2}{169})$

so fractional loss in energy is given as

$f = \frac{\Delta K}{K}$

$f = \frac{\frac{60}{169}mv_o^2}{\frac{1}{2}mv_o^2}$

$f = \frac{120}{169}$

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

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Explanation:

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