All categories

3 years ago

If an object of mass 70kg falls from a height of 500 m, what is the maximum velocity of the object?

Physics

1 answer:

MatroZZZ [7]3 years ago

5 0

Answer:

We can use this expression to find the maximum velocity or velocity just before touching the ground. For this problem, v = √2×10×500 = 100m/s.

You might be interested in

The ratio of output power to input power, in percent, is called?

Harrizon [31]

That's efficiency. There's no law that it must be stated in percent.

4 0

3 years ago

Read 2 more answers

in a cricket match there are 5000 spectators counted 10 by 10 the number of significant figure in the measurement will be

BabaBlast [244]

Answer:

$\boxed{3 \ Significant \ figures}$

Explanation:

Total spectators = 5000

Counted by the groups of ten, So at last the result will be:

=> 5000/10 = 500

Significant figures in 500 are 3

8 0

3 years ago

According to the law of mass, 8 grams of hydrogen and 32 grams of oxygen will combine to form how many grams of water?

seropon [69]

Answer:

36 grams of water

Explanation:

mass of H2O = 0.5 x 18 = 9g

7 0

2 years ago

A. What is the coefficient of kinetic friction?

Andrews [41]

Explanation:

Coefficient of kinetic friction is the resistive force that opposes the motion of a body as it moves and is in contact with another body.

It is found by dividing the frictional force by the normal force.

Friction is a force that opposes motion.
Static friction is for bodies that are not in motion
Kinetic friction is for moving bodies.

7 0

3 years ago

A 12.0-g bullet is fi red horizontally into a 100-g wooden block initially at rest on a horizontal surface. After impact, the bl

allochka39001 [22]

Answer:

$v_1 = 91.3 m/s$

Explanation:

By energy conservation and work energy theorem we can say that after bullet hits the block, it will move on the rough floor and comes to rest

so here work done by frictional force = change in kinetic energy

so we know that

$W_f = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

$-\mu mg d = \frac{1}{2}m(v_f^2 - v_i^2)$

$-0.650(9.81)(7.5) = \frac{1}{2}(0 - v_i^2)$

$47.8 = \frac{1}{2}v^2$

$v = 9.78 m/s$

now by momentum conservation we have

$m_1 v_1 = (m_1 + m_2) v$

$12 v_1 = (100 + 12) 9.78$

$v_1 = 91.3 m/s$

3 0

2 years ago