An object is placed at a distance 30cm from a thin lenses of power 4 diaptor. Discuss the nature of image

Korolek [52]

Hi how are you?

Are you a Libra?

8 0

3 years ago

A ball rolls down the hill which has a vertical height of 15 m. Ignoring friction what would be the gravitational potential ener

trasher [3.6K]

a) Potential energy: 147 m [J]

The gravitational potential energy of an object is given by

$U=mgh$

where

m is its mass

$g=9.8 m/s^2$ is the acceleration of gravity

h is the height of the object above the ground

In this problem,

h = 15 m

We call 'm' the mass of the ball, since we don't know it

So, the potential energy of the ball at the top of the hill is

$U=(m)(9.8)(15)=147 m$ (J)

b) Velocity of the ball at the bottom of the hill: 17.1 m/s

According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:

$U=K=\frac{1}{2}mv^2$

where

v is the final velocity of the ball

We know from part a) that

U = 147 m

Substituting into the equation above,

$147 m = \frac{1}{2}mv^2$

And re-arranging for v, we find the velocity:

$v=\sqrt{2\cdot 147}=17.1 m/s$

5 0

4 years ago

Which one?? please someone quick!

LiRa [457]

Answer:

i think its second law of motion.

Explanation:

4 0

3 years ago

Read 2 more answers

21) A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest

zlopas [31]

Answer:

-22,150 N

Explanation:

When the youngster jumps off the platform, during the fall her initial potential energy is converted into kinetic energy, according to the law of conservation of energy. Therefore, we can write:

$mgh=\frac{1}{2}mu^2$