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cricket20 [7]
3 years ago
5

A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same di

rection at another constant speed. the average velocity for the entire trip is 26.5 mi/h. (a) what is the constant speed with which the car moved during the second distance d?
Physics
1 answer:
Shalnov [3]3 years ago
5 0

A distance of d is covered with 53 mile/hr initially. Time taken to cover this distance t1 = d/53 hour Next distance of d is covered with x mile hours. Time taken to cover this distance t2 = d/x hours. We have average speed = 26.5 mile / hour          

                                         = Total distance traveled/ total time taken                  

                                         = \frac{2d}{\frac{d}{53}+\frac{d}{x}} = \frac{2}{\frac{1}{53}+\frac{1}{x} }  = \frac{106x}{x+53}

                              26.5 = \frac{106x}{x+53} \\ \\ 79.5 x = 1404.5\\ \\ x = 17.67 miles/hour

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Gekata [30.6K]

The specific heat capacity of the block is 508J/kg^{\circ}C

Explanation:

As the block is placed into the water, heat energy is transferred from the water (which is at higher temperature) to the block (which is at lower temperature), until the block and the water are in thermal equilibrium (= same temperature).

Therefore, we can write:

Q_{water}=Q_{block}

Where

Q_{water}=m_w C_w (T_w-T_{eq}) is the heat energy released by the water, where

m_w = 0.217 kg is the mass of the water

C_w = 4186 J/kg^{\circ}C is the water heat specific capacity

T_w = 25.0^{\circ} is the initial temperature of the water

T_{eq}=16.4^{\circ} is the temperature at equilibrium

Substituting,

Q_{water}=(0.217)(4186)(25.0-16.4)=7812 J

Now we can write the heat energy absorbed by the block as

Q_{block}=m_b C_b(T_{eq}-T_b)

where

m_b=0.350 kg is the mass of the block

C_b is the specific heat capacity of the block

T_b = -27.5^{\circ} is the initial temperature of the block

And solving for C_b,

C_b=\frac{Q_{block}}{m_b(T_{eq}-T_b)}=\frac{7812}{(0.350)(16.4-(-27.5))}=508J/kg^{\circ}C

Learn more about specific heat capacity:

brainly.com/question/3032746

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Answer:

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Work is force times distance. So here we have
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