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3 years ago

Weight of a man is less at coal mine.why?

1 answer:

7 0

If he is underground then there is less planetary mass below him and therefore less gravity. weight is just gravity's affect on you. his mass won't change.

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Why don't you notice your gravitational force on other objects?

mixas84 [53]

The answer is B tell me if I am wrong.

4 0

3 years ago

Read 2 more answers

A super ball stops bouncing because A) gravity never lets it bounce. B) it loses energy due to friction. C) it cannot gain poten

AlexFokin [52]

Answer:

Explanation:

it loses energy due to friction because if it hits objects the objects slow it down and which causes it to start slowing down

6 0

3 years ago

A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta

Vika [28.1K]

Answer:

$t_t=4.131\ s$

Explanation:

Given:

height above the horizontal form where the ball is hit, $y=1\ m$

angle of projectile above the horizontal, $\theta=30^{\circ}$

initial speed of the projectile, $u=40\ m.s^{-1}$

Firstly we find the vertical component of the initial velocity:

$u_y=u.\sin\theta$

$u_y=40\times \sin30^{\circ}$

$u_y=20\ m.s^{-1}$

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

$v_y=0\ m.s^{-1}$

Using eq. of motion:

$v_y^2=u_y^2-2g.h$ (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

$0^2=20^2-2\times 9.8\times h$

$h=20.4082\ m$ (from the height where it is thrown)

Now we find the time taken to ascend to this height:

$v_y=u_y-g.t$

$0=20-9.8t$

$t=2.041\ s$

Time taken to descent the total height:

we've total height, $h'=h+y$ $=20.4082+1$

$h'=u_y'.t'+\frac{1}{2} g.t'^2$

during the course of descend its initial vertical velocity is zero because it is at the top height, so $u_y'=0\ m.s^{-1}$

$21.4082=0+4.9t'^2$

$t'=2.09\ s$

Now the total time taken by the ball to hit the ground:

$t_t=t'+t$

$t_t=2.09+2.041$

$t_t=4.131\ s$

3 0

3 years ago

The tomato is dropped. What is the velocity, v, of the tomato when it hits the ground? Assume 85.6 % of the work done in Part A

IRISSAK [1]

Answer:

$v = 4.1 \sqrt{h}$

Explanation:

Let the mass of tomato is m and the height from which it falls is h.

Let the tomato its the ground with velocity v.

The potential energy of the tomato at height h

U = m x g x h

The kinetic energy of tomato as it hits the ground

K = 1/2 mv^2

According to the question,

85.6 % of Potential energy = Kinetic energy

$\frac{85.6}{100}\times m\times g\times h = \frac{1}{2}\times m\times v^{2}$

$v = 4.1 \sqrt{h}$

7 0

4 years ago

Read 2 more answers

A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the ca

igomit [66]

Answer:

Force of friction, f = 751.97 N

Explanation:

it is given that,

Mass of the car, m = 1100 kg

It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.

From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.

$f=mg\ sin\theta$

$f=1100\ kg\times 9.8\ m/s^2\ sin(4)$

f = 751.97 N

So, the force of friction on the car is 751.97 N. Hence, this is the required solution.

3 0

3 years ago