Which term describes energy stored inside the nucleus of an atom?

Write an essay about the things an individual must do in order to adapt and survive the changes that are happening in his/her en

Scorpion4ik [409]

Sorry Man If you ask this question no one wants to write an essay and waste their time. You are smart enough to write an essay.

7 0

2 years ago

A 45 N girl sits on a bench 0.6 meters off the ground. How much work is done on the bench?

ycow [4]

Answer: 27 joules

Explanation:

Work is done when force is applied on the bench over a distance. it is measured in joules.

Workdone = force x distance

= 45 N x 0.6 metres

= 27 joules

Thus, 27 joules of work is done on the bench.

6 0

3 years ago

If element X has 40 protons will it also have 40 electrons

Nataliya [291]

Yes, atoms need to be equal in protons and electrons

4 0

3 years ago

Read 2 more answers

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400

liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, $f_{ra} = \mu mg\\$ ............(1)

Since frictional force is a type of force, we are safe to say $f_{ra} = ma$ .......(2)

Equating (1) and (2)

$ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}$

a) Speed of A at the start of the sliding

$d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s$

b) Speed of B at the start of the sliding

$d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s$