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Viktor [21]
3 years ago
10

A conducting rod moves perpendicularly through a uniform magnetic field. How does the emf change if the magnetic field is increa

sed by a factor of 4 and the time required for the rod to move through the field is decreased to half of the original time?
a) It increases by a factor of 2
b) It increases by a factor of 4
c) It increases by a factor of 8
d) It increases by a factor of 16 .
Physics
1 answer:
Marysya12 [62]3 years ago
5 0

Answer:

c) It increases by a factor of 8

Explanation:

According to Faraday's law (and Lenz' law), the induced EMF is given as the rate of change of magnetic flux.

Mathematically:

V = -dФ/dt

Magnetic flux, Ф, is given as:

Ф = BA

where B  = magnetic field strength and A = Area of object

Hence, induced EMF becomes:

V = -d(BA)/dt  or -BA/t

If the magnetic field is increased by a factor of 4, (B_n = 4B) and the time required for the rod to move is decreased by a factor of 2 (t_n = t/2), the induced EMF becomes:

V_n = -(B_nA)/t_n

V_n = \frac{-4BA}{(t/2)}\\\\V_n = \frac{-8BA}{t} \\\\V_n = 8V\\

The EMF has increased by a factor of 8.

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A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d
Iteru [2.4K]

Answer:

A point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

                         

Explanation:

We can find the distance with the following equation since the acceleration is cero (the disk rotates at a constant rate):

d = v*t

Where:

v: is the tangential speed of the disk

t: is the time = 30 s  

The tangential speed can be found as follows:

v = \omega*r

Where:

ω: is the angular speed = 100 rpm

r: is the radius = 50 cm = 0.50 m

v = \omega*r = 100 \frac{rev}{min}*\frac{2\pi rad}{1 rev}*\frac{1 min}{60 s}*0.50 m = 5.24 m/s    

Now, the distance traveled by the disk is:

d = v*t = 5.24 m/s*30 s = 157.2 m

Therefore, a point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

I hope it helps you!

3 0
2 years ago
Estimate frequency of vibration of your arm. Let the length of the arm be 0.57 m. Consider the arm as a simple pendulum and assu
skad [1K]

Answer:

0.80865 Hz

1.23662 seconds

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

l = Length of arm = 0.57 m

Length of simple pendulum is given by

L=\dfrac{2}{3}l\\\Rightarrow L=\dfrac{2}{3}\times 0.57\\\Rightarrow L=0.38\ m

The frequency is given by

f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}}\\\Rightarrow f=\dfrac{1}{2\pi}\sqrt{\dfrac{9.81}{0.38}}\\\Rightarrow f=0.80865\ Hz

The frequency is 0.80865 Hz

The time period is given by

T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{0.80865}\\\Rightarrow T=1.23662\ s

The time period is 1.23662 seconds

3 0
2 years ago
In this experiment, you will use a track, a toy car, and some washers to explore Newton’s first two laws of motion. You will mak
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How can we experimentally verify newton's laws?

7 0
3 years ago
Read 2 more answers
A plastic film moves over two drums. During a 4-s interval the speed of the tape is increased uniformly from v0 = 2ft/s to v1 =
Ratling [72]

Answer:

Question 1)

a) The speed of the drums is increased from 2 ft/s to 4 ft/s in 4 s. From the below kinematic equations the acceleration of the drums can be determined.

v_1 = v_0 + at \\4 = 2 + 4a\\a = 0.5~ft/s^2

This is the linear acceleration of the drums. Since the tape does not slip on the drums, by the rule of rolling without slipping,

v = \omega R\\a = \alpha R

where α is the angular acceleration.

In order to continue this question, the radius of the drums should be given.

Let us denote the radius of the drums as R, the angular acceleration of drum B is

α = 0.5/R.

b) The distance travelled by the drums can be found by the following kinematics formula:

v_1^2 = v_0^2 + 2ax\\4^2 = 2^2 + 2(0.5)x\\x = 12 ft

One revolution is equal to the circumference of the drum. So, total number of revolutions is

x / (2\pi R) = 6/(\pi R)

Question 2)

a) In a rocket propulsion question, the acceleration of the rocket can be found by the following formula:

a = \frac{dv}{dt} = -\frac{v_{fuel}}{m}\frac{dm}{dt} = -\frac{13000}{2600}25 = 125~ft/s^2

b) a = -\frac{v_{fuel}}{m}\frac{dm}{dt} = - \frac{13000}{400}25 = 812.5~ft/s^2

5 0
3 years ago
Which are most likely to be used during a routine visit to the dentist?
Paraphin [41]

Answer:

The most likely items to be used are;

Ultrasound and X-rays

Explanation:

A routine visit to a dentist consists of two areas of activities, including;

a) Dental examination and check up

b) Oral prophylaxis, and dental cleaning

The dental examination may involve the use of X-rays, which allow the detection of cavities between the teeth

The dental cleaning can be carried out with the use of an ultrasound cleaner, which allow the cleaning of sensitive teeth without hurting the patient

Therefore, the items most likely to be used during a routine dental visit are ultrasound and X-rays

8 0
2 years ago
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