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3 years ago

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In a crash test, a 1000 kg automobile moving at 10 m/s crashes into a brick wall. How much energy goes into demolishing and warm

ing the wall and the auto?

1 answer:

marishachu [46]3 years ago

8 0

Answer:

Explanation:

Given

mass of automobile $m=1000\ kg$

velocity of automobile $v=10\ m/s$

If the automobile crashes into the brick wall then it's kinetic energy is converted into thermal and demolishing energy to wall.

Car completely stops as it crashes into the wall so Energy converted into demolition and thermal energy

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

$E=\frac{1}{2}\times 1000\times (10)^2$

$E=50\ kJ$

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3. A concrete highway is built of slabs 12 m long (20o C). How wide should the expansion cracks between the slabs be (at 20o C)

lesantik [10]

Answer:

$\Delta x=2.304\times 10^{-2}\ m=2.304\ cm$ is the minimum gap between the slabs

Explanation:

Given:

length of the concrete highway, $l=12\ m$

coefficient of thermal expansion, $\alpha=12\times 10^{-6}\ ^{\circ}C^{-1}$

range of temperature variation, $(-30^{\circ}C\ to\ 50^{\circ}C) \Rightarrow \Delta T=80^{\circ}C$

<u>Now form the equation of thermal expansion:</u>

$\Delta l=l\times \alpha\times \Delta T$

$\Delta l=12\times (12\times 10^{-6})\times 80$

$\Delta l=1.152\times 10^{-2}\ m$

Since each slab of the highway expands by the above length so the minimum gap between the slabs to prevent buckling:

$\Delta x=2\times \Delta l$

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In a wire, when elongation is 4 cm energy stored is E. if it is stretched by 4 cm, then what amount of elastic potential energy

myrzilka [38]

<h2>Answer:</h2>

4E

<h2>Explanation:</h2>

The elastic potential energy of an elastic material (e.g a spring, a wire), is the energy stored when the material is stretched or compressed. It is given by

U = $\frac{1}{2}kx^2$ --------------------(i)

Where;

U = potential energy stored

k = spring constant of the material

x = elongation (extension or compression of the material).

<em>From the first statement;</em>

<em>when elongation (x) is 4cm, energy stored (U) is E</em>

<em>Substitute these values into equation (i) as follows;</em>

E = $\frac{1}{2}k(4)^2$

E = 8k

<em>Make k subject of the formula</em>

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<em>It is stretched by 4cm.</em>

This means that total elongation will be 4cm + 4cm = 8cm.

The potential energy stored will be found by substituting the value of x = 8cm and k = $\frac{E}{8}$ into equation (i) as follows;

U = $\frac{1}{2}\frac{E}{8} (8)^2$

U = $\frac{1}{2}{8E}$

U = ${4E}$

Therefore, the potential energy stored will now be 4 times the original one.

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