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jeka57 [31]
2 years ago
7

A copper sphere was moving at 25 m/s when it hit another object. This caused all of the KE to be converted into thermal energy f

or the copper sphere. If the specific heat capacity of copper is 387 J/(kg ⋅ C°), what was the increase in temperature?
0.23 C°
0.81 C°
1.3 C°
2.1 C°
Physics
1 answer:
Verdich [7]2 years ago
3 0

Answer:

\Delta T = 0.81 ^oC

Explanation:

As we know by energy conservation

All its kinetic energy will convert into thermal energy to raise its temperature

\frac{1}{2}mv^2 = ms\Delta T

now divide both sides by mass of the object

\frac{1}{2}v^2 = s\Delta T

so change in temperature is given as

\Delta T = \frac{v^2}{2s}

\Delta T = \frac{25^2}{2\times 387}

\Delta T = 0.81^oC

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For flowing water, what is the magnitude of the velocity gradient needed to produce a shear stress of 1.0 n/m2 ?
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Two objects have masses m and 5m, respectively. They both are placed side by side on a frictionless inclined plane and allowed t
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Answer:

(E) The two objects reach the bottom of the incline at the same time.

Explanation:

Given;

first object with mass, m

second object with mass, 5m

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Since both objects have the same acceleration of gravity, and no external force due friction (frictionless inclined plane), they will reach bottom of the inclined at the time.

Thus, the acceleration due to gravity is constant for all objects regardless of their masses.

Therefore, the correct option is E;

(E) The two objects reach the bottom of the incline at the same time.

5 0
2 years ago
a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres
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Answer:

Part a) When collision is perfectly inelastic

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b) When collision is perfectly elastic

v_m = \frac{m + M}{2m}\sqrt{5Rg}

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

mv_m = (m + M)v

so we have

v = \frac{mv_m}{m + M}

now we know that in order to complete the circle we will have

v = \sqrt{5Rg}

\frac{mv_m}{m + M} = \sqrt{5Rg}

now we have

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b)

Now we know that collision is perfectly elastic

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v = \frac{2mv_m}{m + M}

now we have

\sqrt{5Rg} = \frac{2mv_m}{m + M}

v_m = \frac{m + M}{2m}\sqrt{5Rg}

6 0
2 years ago
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When some amount of mass is lost, same amount of energy equivalent to mass is produced.

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The energy produced in this reaction is nothing else than the energy equivalent to mass defect.  Approximately 199.5 Mev of energy equivalent to this mass defect is produced in this reaction.

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