If Car A has a greater momentum than Car B, what will happen to both vehicles immediately after they collide?

A thin Nichrome wire connected to an ammeter surrounds a region of time-varying magnetic flux, and the ammeter reads 13 amperes.

Semenov [28]

Answer:

The current would be same in both situation.

Explanation:

Given that,

Current I = 13 A

Number of turns = 23

We need to calculate the induced emf

Using formula of induced emf is

$\epsilon=NA\dfrac{dB}{dt}$

For N = 1

$\epsilon=A\dfrac{dB}{dt}$

We need to calculate the current

Using formula of current

$i=\dfrac{\epsilon}{R}$

Put the value of emf

$i=\dfrac{A\dfrac{dB}{dt}}{R}$

Now, if the number of turn is 22 , then induced emf would be

$\epsilon'=NA\dfrac{dB}{dt}$

Then the current would be

$i'=\dfrac{\epsilon'}{NR}$

$i'=\dfrac{NA\dfrac{dB}{dt}}{NR}$

$i'=\dfrac{A\dfrac{dB}{dt}}{R}$

$i'=i$

Hence, The current would be same in both situation.

4 0

3 years ago

Which of the following happens to an object in uniform circular motion?

podryga [215]

Answer:

As an object moves in a circle, it is constantly changing its direction. ... Accelerating objects are objects which are changing their velocity - either the speed (i.e., magnitude of the velocity vector) or the direction. An object undergoing uniform circular motion is moving with a constant speed.

5 0

3 years ago

What force must act on a 50.0-kg mass to give it an ancceleration of 0.30 m/s^2?

ratelena [41]

Answer:

15.0 N

Explanation:

see pic

3 0

2 years ago

A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with

Andrej [43]

Answer:

5.4 J.

Explanation:

Given,

mass of the object, m = 2 Kg

initial speed, u = 5 m/s

mass of another object,m' = 3 kg

initial speed of another orbit,u' = 2 m/s

KE lost after collusion = ?

Final velocity of the system

Using conservation of momentum

m u + m'u' = (m + m') V

2 x 5 + 3 x 2 = ( 2 + 3 )V

16 = 5 V

V = 3.2 m/s

Initial KE = $\dfrac{1}{2}mu^2 + \dfrac{1}{2}m'u'^2$

= $\dfrac{1}{2}\times 2\times 5^2 + \dfrac{1}{2}\times 3 \times 2^2$

= 31 J

Final KE = $\dfrac{1}{2} (m+m')V^2 = \dfrac{1}{2}\times 5 \times 3.2^2 = 25.6 J$

Loss in KE = 31 J - 25.6 J = 5.4 J.

4 0

3 years ago

1. A diver dives off of a raft - what happens to the diver? The raft? How does this relate to Newton's Third Law? Action Force:

Natali [406]

The Action Force of this scenario is the pushing force of the Diver. The Reaction Force is the raft pushing back on the diver.

The Third Law of Motion states that "For every action, there is an equal and opposite reaction." Now when the diver dives off the raft, the raft is also pushing the same amount of force as the diver did as he dives off. The diver will then move forward and the raft on the other hand will move backwards.

The movement of the raft shows the opposite force. It will move backwards depending on how strong the diver will push off on the raft. And the amount of force he pushes on it, the raft will exert the same force so the stronger the force of the diver, the farther he will go because the raft will push him in that same direction as it goes backwards.

7 0

3 years ago