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3 years ago

A 200 kg chunck of lead falls from a hight of 30m and smashes into a rigid floor. calculate the increse in internal engery

Physics

1 answer:

mr Goodwill [35]3 years ago

5 0

Answer:

$\Delta E=58800\ J$

Explanation:

Given:

mass of lead chunk, $m=200\ kg$

height of the fall, $h= 30\ m$

Assuming the collision to be perfectly inelastic the whole kinetic energy of the chunk is lost after the collision.

Here when the chunk is at the given height it has a potential energy which on falling transforms into kinetic energy of the chunk.

So, increase in the internal energy of the system after the collision is :

$\Delta E=m.g.h$

$\Delta E=200\times 9.8\times 30$

$\Delta E=58800\ J$

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3 years ago

A weight lifter lifts a dumbbell a certain height in 2.0 s, while a competitor does the same workin 1.0 s. Compared to the power

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Answer:

a. one-half as great

Explanation:

The power developed by the first lifter is one-half as great as that of the second person.

Power is defined as the rate at which work is done;

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Since the two lifters do the same work at different time, let us estimate their power;

P₁ = $\frac{workdone}{2}$ P₂ = $\frac{workdone }{1}$

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3 years ago

A 2 m tall, 0.5 m inside diameter tank is filled with water. A 10 cm hole is opened 0.75 m from the bottom of the tank. What is

valina [46]

Answer:

4.75 m/s

Explanation:

The computation of the velocity of the existing water is shown below:

Data provided in the question

Tall = 2 m

Inside diameter tank = 2m

Hole opened = 10 cm

Bottom of the tank = 0.75 m

Based on the above information, first we have to determine the height which is

= 2 - 0.75 - 0.10

= 2 - 0.85

= 1.15 m

We assume the following things

1. Compressible flow

2. Stream line followed

Now applied the Bernoulli equation to section 1 and 2

So we get

$\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2$

where,

P_1 = P_2 = hydrostatic

z_1 = 0

z_2 = h

Now

$\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\ \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}$

= 4.7476 m/sec

= 4.75 m/s

6 0

3 years ago