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hjlf
3 years ago
8

When a bar magnet is thrust at the same speed into a coil having twice the number of loops, the induced voltage is the same, no

different. half. twice as much. four times as much. none of the above
Physics
1 answer:
ss7ja [257]3 years ago
6 0

Answer:

Twice as much

Explanation:

Changing magnetic flux through a coil produce emf in the coil.

According to Faraday's law induced voltage is given by

∈= - N *dФ \ dt -----------( 1 )

N is number of turns of coil

Ф is flux linkage

dФ/dt is rate of change of flux

From equation (1) it is clear if rate of change of flux is the same but we  doubled the number of loops (i.e N₂= 2N₁) the voltage induced will be double.

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The chart shows data for a moving object.
Brut [27]
The object is not accelerating
8 0
3 years ago
An 800-N billboard worker stands on a 4.0-m scaffold supported by vertical ropes at each end. If the scaffold weighs 500 N and t
Arturiano [62]

Answer:

T = 850 N

Explanation:

given,

mass of billboard worker = 800 N

length of scaffold = 4 m

weight of the scaffold = 500 N

worker is standing at 1 m from one end.

Tension in the rope = ?

To calculate the tension in the string we have to balance the clockwise and counterclockwise moment of the system.

Weight of the worker and the weight of the scaffold will be in clockwise direction where as the tension will be in counterclockwise direction

now,

800 x 3 + 500 x 2 = T x 4

4T = 3400

T = 850 N

hence, tension in the rope is equal to 850 N

5 0
4 years ago
10. A boy weighs 475 N. What is his mass? (acceleration due to gravity on Earth is 9.8m/s2 = g)
Darya [45]

Answer: mass = 48.47 kg.

Explanation:

Formula : Weight = mg  , where m = mass of body , g= acceleration due to gravity .

Given: Weight  = 475 N

g= 9.8\ m/s^2

Substitute all values in formula ,  we get

475= m \times9.8\\\\\Rightarrow\ m = \dfrac{475}{9.8}\\\\\Rightarrow\ m \approx 48.47\ kg

Hence, his mass = 48.47 kg.

7 0
3 years ago
The earth’s radius is 6.37 × 10 6 m ; it rotates once every 24 hours. What is the earth’s angular speed? Viewed from a point abo
Advocard [28]

Answer:

a) the angular speed of the Earth's rotation is <em>7.272 × 10⁻⁵ rad/s.</em>

<em></em>

b) Viewed from the North Pole, Earth's angular speed rotates in an anticlockwise direction. Therefore, the <em>angular velocity is positive.</em>

<em></em>

c) Earth's speed at a point on the equator is <em>463.23 m/s.</em>

<em></em>

d)  the speed of a point on the earth’s surface halfway between the equator and the pole is <em>231.61 m/s.</em>

Explanation:

a) The angular speed of the Earth's rotation is:

ω = 2π / T

where

T is the period

ω = 2π / (24 hr × (3600 s / 1 hr))

<em>ω = 7.272 × 10⁻⁵ rad/s</em>

b) Viewed from the North Pole, Earth's angular speed rotates in an anticlockwise direction. Therefore, the <em>angular velocity is positive.</em>

c) Earth's speed at a point on the equator is:

v = r × ω

v = (6.37 × 10⁶ m) × (7.272 × 10⁻⁵ rad/s)

<em>v = 463.23 m/s</em>

d) The radius of the circle in which the point moves is half of Earth's radius. Therefore,

r = 1/2(6.37 × 10⁶ m)

r = 3.19 × 10⁶ m

Therefore, the speed of a point on the earth’s surface halfway between the equator and the pole is:

v = r × ω

v = (3.19 × 10⁶ m) × (7.272 × 10⁻⁵ rad/s)

<em>v = 231.61 m/s</em>

6 0
3 years ago
What is the specific heat of the masses in this experiment? Infer the substance the masses are made of and explain your inferenc
Liono4ka [1.6K]

The metal whose specific heat capacity is close to the obtained value is aluminum.

<h3>What is specific heat capacity</h3>

The specific heat capacity of an object is the heat required to raise a unit mass of the substance by 1 kelvin.

Q= mc\Delta \theta

where;

  • c is the specific heat capacity
  • Δθ is change in temperature

Let the mass of the water = 50 g

mass of the metal for this first trial = 50 g

The heat gained by the water is calculated as follows

Q = 50 \times 4.184 \times 8.4\\\\Q = 1757.28 \ J

Specific heat capacity of the metal for the first trial is calculated as follows;

Heat gained by water = Heat lost by metal

C = \frac{Q}{m\Delta T} = \frac{1757.28}{50\times 8.4} = 4.184 \ J/g^oC

Specific heat capacity of the metal for the second trial;

mass of metal = 200 - mass of water = 150 g

C_2 = \frac{1757.28}{150 \times 15.2} = 0.77 \ J/g^oC

Specific heat capacity of the metal for the third trial;

C_3 = \frac{1757.28}{250 \times 20.8} = 0.34\ J/g^oC

Specific heat capacity of the metal for the fourth trial;

C_4 = \frac{1757.28}{350 \times 25.4} = 0.19\ J/g^oC

Specific heat capacity of the metal for the fifth trial;

C_5 = \frac{1757.28}{450 \times 29.6} = 0.13\ J/g^oC

Average specific heat capacity

C = \frac{4.184 + 0.77 + 0.34+ 0.19 + 0.13 }{5} = 1.12 \ J/g^oC = 1120 J/kg^oC

The metal whose specific heat capacity is close to the above value is aluminum.

Learn more about specific heat capacity here: brainly.com/question/16559442

8 0
2 years ago
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