All categories

3 years ago

5. i) Name two devices based on interaction between magnetic field and current carrying conductor.

Physics

1 answer:

anyanavicka [17]3 years ago

5 0

Answer:

some common devices that use current carrying conductors and magnetic fields are electric motor electric generator loudspeakers microphones and measuring instruments like galvanometer ammeter and voltmeter

You might be interested in

A truck traveling at a constant speed of 28 m/s passes a more slowly moving car. The instant the truck passes the car, the car b

choli [55]

Answer:

the velocity of car when it passes the truck is u = 16.33 m/s

Explanation:

given,

constant speed of truck = 28 m/s

acceleration of car = 1.2 m/s²

passes the truck in 545 m

speed of the car when it just pass the truck = ?

time taken by the truck to travel 545 m

time = $\dfrac{distance}{speed}$

time = $\dfrac{545}{28}$

time =19.46 s

velocity of the car when it crosses the truck

$S = ut + \dfrac{1}{2}at^2$

$545= u\times 19.46 + \dfrac{1}{2} \times 1.2 \times 19.46^2$

u = 16.33 m/s

the velocity of car when it passes the truck is u = 16.33 m/s

5 0

4 years ago

A scaffold of mass 77 kg and length 5.0 m is supported in a horizontal position by a vertical cable at each end. A window washer

Lynna [10]

Answer:

A) T2 = 912.88 N

B) T1 = 607.12 N

Explanation:

First of all, we see that the sum of the tensions is equal to the total weight.

Now, for the scaffold, weight; W_s = 77 x 9.8 = 755.6 N

For the window washer, Weight; W_w = 78 x 9.8 = 764.4 N

Total weight;W_t = W_s + W_w

W_t = 755.6 N + 764.4 N = 1520 N

Thus,

T1 + T2 = 1520

Where T1 and T2 are the tensions in farther and nearer cables respectively.

Now, we need to do a torque problem.

The window washer is 1.8m from the right end of the scaffold and so the weight of the scaffold is at its center. This is 2.5 m from either end. Let the pivot point be at right end of the scaffold.

For the window washer, counter clockwise torque = 764.4 x 1.5 = 1146.6 N.m

For the scaffold, counter clockwise torque = 755.6 x 2.5 = 1889 N.m

Total Torque; T = 1146.6 + 1889 = 3035.6 N.m

For the cable at the left end of the cable, clockwise torque = T1 x 5

Set this equal to the total counter clockwise torque and solve for T1.

Thus,

T1 x 5 = 3035.6

T1 = 3035.6 ÷ 5 = 607.12 N

T2 = 1520 – 607.12 = 912.88 N

6 0

4 years ago

I rent a small high pressure water sprayer to clean the outside of my house. The sprayer works like a super soaker with a hose b

viva [34]

Answer:

1. 80,000 Pa

2. 11.3 m/s

3. 12.5 m/s

Explanation:

Question 1

Pressure, $P=hg\rho$

Where h is the height that water is to reach, g is gravitational constant and $\rho$ is the density, in this case, we assume $\rho$ of pure water as $1000 Kg/m^3$

Assuming $g=10 m/s^{2}$

P=8*10*1000=80000 Pa

Question 2

Pressure can also be found by the formula

$P=0.5v^{2}\rho$ where v is the velocity

Equating the new formula of pressure to the formula used in question 1 above

$P=0.5v^{2}\rho=hg\rho$

Notice that $\rho$ is common hence

$0.5v^{2}=hg$

Making V the subject of the formula

$v^{2}=2hg$

$v=\sqrt 2hg$

In this case, h=8-1.6=6.4m and taking g as 10 m/s^{2}

$v=\sqrt 2*10*6.4=11.3137085 m/s$

Rounding off to 1 decimal place

v=11.3 m/s

Question 3

As already illustrated

$v=\sqrt 2hg$

Taking g as 9.8 and h now is 8m

$v=\sqrt 2*8*9.8$

v=12.52198067

Rounding off to 1 decimal place

v=12.5 m/s

6 0

3 years ago

An alert physics student stands beside the tracks as a train rolls slowly past. He notes that the frequency of the train whistle

Dovator [93]

Answer:The velocity of the train is 3.84m/s

Explanation:

According to the Doppler effect, if the source is moving towards you then the apparent frequency of the sound emitted by the source is higher and if the source is moving away from you then the apparent frequency of the sound emitted by the source is smaller.

This is given by:

fo = V +-Vo/ V +-Vo × source

Where fo= observed frequency

V= velocity of sound

Vo= vo it of the observer

fsource= frequency the source

Given:

Observed frequency of the approaching train fo1= 452Hz

The observed frequency of train= fo2= 442Hz

Velocity of sound= 334m/s

Velocity of source=?

Train approaching the observer is given by:

fo1= V/(V - Vs)× source ...eq1

Train passes the student is given by:

fo2= V/(V - Vs)×source ...eq2

Divide eq1 by eq2

452/442 = (343+Vs)/(343 - Vs)

1.02 =(343+Vs)/(343 -Vs)

Cross multiply

1.02(343- Vs) = 343 + Vs

350.76 - 1.02Vs = 343 + Vs

Collecting like terms

350.76 -343= 1.02Vs+ Vs

7.76 = 2.02Vs

Vs= 7.76/2.02

Vs= 3.84m/s

4 0

3 years ago

3. A snowmobile has a mass of 2.25 x 102 kg. A constant force is exerted on it for 50.0 s. The snowmobile’s initial velocity is

Art [367]

Answer:

a. m(v-u)

momentum =5175kgm/s

b. 103.5N

3 0

3 years ago