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2 years ago

Match each situation

Physics

1 answer:

EastWind [94]2 years ago

3 0

A: is potential
C: is losing kinetic energy and gaining potential energy
B: kinetic energy is at its highest
D: is loosing potential energy and gaining kinetic energy

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A 0.45 m radius, 500 turn coil is rotated one-fourth of a revolution in 4.01 ms, originally having its plane perpendicular to a

AysviL [449]

Answer:

B = 0.126 T

Explanation:

As per Faraday's law we know that rate of change in magnetic flux will induce EMF in the coil

So here we can say that EMF induced in the coil is given as

$EMF = \frac{\phi_2 - \phi_1}{\Delta t}$

initially the coil area is perpendicular to the magnetic field

and after one fourth rotation of coil the area vector of coil will be turned by 90 degree

so we can say

$\phi_1 = NBAcos 0 = BA$

$\phi_2 = NBAcos 90 = 0$

now we will have

$EMF = \frac{NBA}{t}$

$10,000 V = \frac{(500)(B)(\pi \times 0.45^2)}{4.01\times 10^{-3}}$

$B = 0.126 T$

3 0

2 years ago

Suppose you have a dipole that's free to move in any way (including rotate - imagine it floating in space). And there's an objec

Setler [38]

Complete Question

The complete question shown on the first uploaded image

Answer:

The force on Q due to dipole is Attractive

The charge Q exerts attractive force on the dipole

Yes from the above parts, force depends on the sign of charge

$F = kQq[\frac{d^{2}+2rd}{r^{2}(d+r)^{2}} ]$

The magnitude o force decrease by a factor of 8.0 times

Explanation:

The explanation is shown on the second uploaded image

5 0

2 years ago

4) Block A has a mass of 3kg and velocity of 13m/s, catching up with a second block B that has a mass of 3kg and is moving with

serious [3.7K]

Answer:

Option A is the correct answer.

Explanation:

Here momentum is conserved.

That is $\left (m_Av_A+m_Bv_B \right )_{initial}=\left (m_Av_A+m_Bv_B \right )_{final}$

Substituting values

$3\times 13+3\times 5=3v_A+3\times 8\\\\3v_A=39+15-24\\\\3v_A=30\\\\v_A=10m/s$

Speed of block A after collision = 10 m/s

Option A is the correct answer.

5 0

2 years ago

A car moves with an initial velocity of 19 m/s due north. (Part A) Find the velocity of the car after 5.6 s if its acceleration

galben [10]

Answer

Assuming

east is the positive x direction

north is the positive y direction

initial velocity , u = 19 j m/s

a) acceleration , a = 1.6 j m/s^2

Using first equation of motion

v = u + a × t

v = 19 + 5.6× 1.6

v = 28 j m/s

the velocity of the car after 5.6 s is 28 m/s north

acceleration , a = -1.5 j m/s^2

Using first equation of motion

v = u + a × t

v = 19 - 5.6 ×1.5

v = 10.6 j m/s

the velocity of the car after 5.6 s is 10.6 m/s north

5 0

2 years ago

If you drop an object from rest, the distance it falls is given by (1/2)at2, where a is the acceleration of the object and t is

klio [65]

If,

$d \: = \: \frac{1}{2} a {t}^{2}$

then, with 3x time t, (suppose, new distance is h)

$h \: = \: \frac{1}{2} a {(3t)}^{2}$

$= \frac{1}{2} a9 {t}^{2}$

$= 9 \: \frac{1}{2} a{t}^{2}$

$= 9d$

Therefore, new distance h will be 9 times bigger than distance d.

answer: c

7 0

2 years ago