A merry-go-round, with mass m and moment

Help needed.

Andreyy89

Answer:

1. -2×10 ^-9

2. 1800v

3 (ii) 4micro Farhad

(iii) V = 3.60 ×10²

(Iv) 45v

Explanation:

I hope it helps

5 0

3 years ago

A sound wave traveling at STP has a period of .0023 seconds. The wavelength of this sound is closest to

agasfer [191]

Answer: 3

Wavelength = 0.76m

Explanation:

When sound wave is travelling at standard temperature and pressure ( STP), its speed will be equal to normal speed of sound which is equal to 331 m/s

Given that the period of the wave is 0.0023 seconds.

Wave speed = wavelength/period

Substitute the wave speed and the period to obtain wavelength

331 = wavelength/0.0023

Cross multiply

Wavelength = 331 × 0.0023

Wavelength = 0.7613 metre

The wavelength of this sound is closest to 0.76 m

4 0

4 years ago

On which moment of inertia of an object does depend on?

Likurg_2 [28]

Additional Information:

I couldn't get your question very clearly. In order to solve the question, I will define moment of inertia, state the formula and factors that the moment of inertia of a body depends and does not depend on.

Answer:

Moment of inertia depends on;

1. Mass of the body

2. Axis of rotation and

3. Distribution of the body

Moment of inertia does not depend on;

1. Angular velocity of the body.

Explanation:

The moment of inertia is defined as a quantity that determines the torque needed for a desired angular acceleration or a property of a body due to which it resists angular acceleration about a rotational axis.

Moment of Inertia, I = ∑mr²

Where,

I is the moment of Inertia

m is the mass

r is the distance from the axis of the rotation

The moment of inertia of a body depends on distribution of body, axis of rotation and mass of the body. However, the moment of Inertia of a body is not dependent on angular velocity of the body.

4 0

4 years ago

A 47.5-turn circular coil of radius 5.25 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0

Free_Kalibri [48]

Answer:

Torque, $5.91\times 10^{-3}\ N-m$

Explanation:

Given that,

The number of turns in the coil, N = 47.5

Radius of the coil, r = 5.25 cm

Uniform magnetic field, B = 0.535 T

Current in the coil, I = 26.9 mA

The magnitude of the maximum possible torque exerted on the coil isg given by :

$\tau=NIAB\\\\\tau=47.5\times 26.9\times 10^{-3}\times \pi (5.25\times 10^{-2})^2\times 0.535 \\\\\tau=0.00591\ N-m\\\\\tau=5.91\times 10^{-3}\ N-m$

So, the magnitude of the maximum possible torque exerted on the coil is $5.91\times 10^{-3}\ N-m$ .

3 0

3 years ago

A 0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of co

Genrish500 [490]

Answer: When 1.0kg of aluminium block is used, the final temperature of the mixture will be T = 36.2∘C

If 1.0kg copper block is used, T of the mixture will be = 17.4∘C

If 100g (0.1kg) of ice at 0∘C is used, T will be = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be= 147.1∘C

Explanation:

H = mcΘ

heat lost by block = heat gained by water

m₁c₁Θ₁ = m₂c₂Θ₂ where m₁ is mass of aluminium, m₂ is mass of water, c₁ is cAluminium, c₂ is cWater, Θ₁ is temperature change for aluminium, Θ₂ is temperature change for water.

0.5*900*(200-20) = m₁*4186*(20-0)

m₁ = 450*180/83270

m₁ = 0.973kg

when 1.0kg of aluminium block is used, the final temperature of the mixture will be T

heat lost by block = heat gained by water

1.0*900*(200-T) = 0.973*4186*(T-0)

180000 - 900T = 4073T

4973T = 180000

T = 180000/4973 = 36.2∘C

If 1.0kg copper block is used, T of the mixture will be

heat lost by block = heat gained by water

1.0*387*(200-T) = 0.973*4186*(T-0)

77400 - 387T = 4073T

4460T = 77400

T = 77400/4460 = 17.4∘C

If 100g (0.1kg) of ice at 0∘C is used, T will be

heat lost by block = heat gained by water + heat used in melting ice to form water at 0∘C

heat used in melting 0.1kg of ice, H = ml, where l= 33600J/Kg

0.5*900*(200-T) = 0.1*4186*(T-0) + 0.1*33600J/Kg

90000 - 450T = 418.6T + 33600

418.6T + 450T = 90000 - 33600

868.6T = 56400

T = 56400/868.6 = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be

0.5*900*(200-T) = 0.025*4186*(T-0) + 0.025*33600J/Kg

90000 - 450T = 104.65T + 8400

104.65T + 450T = 90000 - 8400

554.65T = 81600

T = 81600/554.65 = 147.1∘C

7 0

3 years ago