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olga nikolaevna [1]
3 years ago
5

Tool

Physics
1 answer:
Ede4ka [16]3 years ago
3 0

Answer:

9758 how many significant figures

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Five difference between elastic collision and inelastic collision?​
olga_2 [115]

Answer:

Elastic Collision

Inelastic Collision

The total kinetic energy is conserved. The total kinetic energy of the bodies at the beginning and the end of the collision is different.

Momentum does not change. Momentum changes.

No conversion of energy takes place. Kinetic energy is changed into other energy such as sound or heat energy.

Highly unlikely in the real world as there is almost always a change in energy. This is the normal form of collision in the real world.

An example of this can be swinging balls or a spacecraft flying near a planet but not getting affected by its gravity in the end.

8 0
3 years ago
What is the difference between circular and rotatory motion?​
8090 [49]

Answer:

In a circular motion, the object just moves in a circle. In rotational motion, the object rotates about an axis. ... For example, Earth rotating on its own axis.

3 0
3 years ago
Read 2 more answers
A jogger runs by a river with a velocity of 7 m s relative to the ground. A leaf floating on the river moves with a velocity of
Hunter-Best [27]

Answer:

The velocity of the leaf relative to the jogger is 5 m/s.                    

Explanation:

Given that,

Velocity of jogger wrt to the ground, V_j=7\ m/s

velocity of leaf wrt the ground, v_i=2\ m/s

We need to find the velocity of the leaf relative to the jogger. Let it is equal to V. So, it is given by :

V=v_j-v_i\\\\V=7-2\\\\V=5\ m/s

So, the velocity of the leaf relative to the jogger is 5 m/s. Hence, this is the required solution.

3 0
3 years ago
Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?
Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = \frac{1}{f}

where

f = focal length

Thus

f = \frac{1}{P}

f = \frac{1}{2} = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

7 0
3 years ago
How tightly does mass need to be compacted in order to become a black hole??? (2 words)
olga55 [171]

is it infinite density?

8 0
3 years ago
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