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Katen [24]
2 years ago
11

A string is wrapped around a pulley with a radius of 2.0 cm and no appreciable friction in its axle. The pulley is initially not

turning. A constant force of 50 N is applied to the string, which does not slip, causing the pulley to rotate and the string to unwind. If the string unwinds 1.2 m in 4.9 s, what is the moment of inertia of the pulley?
Physics
2 answers:
stiks02 [169]2 years ago
5 0

Answer:

2

Explanation:

2

sveta [45]2 years ago
5 0

Answer:

Moment of inertia = I = 0.2 kg.m^2

Explanation:

According to equation of motion:

s = ½ at^2

a = 2s/t^2

a = (2 × 1.2)/4.9^2 = 0.09996 m/s^2

The Linear and angular acceleration will be:

a =rα

α = a/r = 0.09996/0.02 = 4.9979 rad/s^2

Torque = T = rF = (0.02)(50) = 1 N

Now,

Moment of inertia = I = T/ α = 1/(4.9979) = 0.2 kg.m^2

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baherus [9]

Answer:

a. Springs oscillate with the same frequency

Explanation:

As they both are in the same height at equilibrium, so

weight of ball must be balanced with spring force, that is

k×x=mg

k= stiffness constant of spring

x=distance stretched

g= acceleration due to gravity

so,  we can write

k/m=g/x

as the g is a constant and they stretched to same distance x so the g/x term becomes constant and

f\propto\sqrt{k/m}

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hence option a is correct.

3 0
2 years ago
In a house the temperature at the surface of a window is 28.9 °C. The temperature outside at the window surface is 7.89 °C. Heat
Alenkasestr [34]

Answer:

-13.18°C

Explanation:

To develop the problem it is necessary to consider the concepts related to the thermal conduction rate.

Its definition is given by the function

\frac{Q}{t} = \frac{kA\Delta T}{d}

Where,

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t = time

k = Thermal conductivity constant

A = Cross-sectional area

\Delta T = The difference in temperature between one side of the material and the other

d= thickness of the material

The problem says that there is a loss of heat twice that of the initial state, that is

Q_2 = 2*Q_1

Replacing,

kA\frac{\Delta T_m}{x} = 2*kA\frac{\Delta T}{x}

\frac{\Delta T}{x}=2*\frac{\Delta T}{x}

\frac{T_i-T_o}{x} = 2\frac{T_1-T_2}{x}

\frac{28.9-T_o}{x} = 2\frac{28.9-7.86}{x}

Solvinf for T_o,

T_o = -13.18

Therefore the temprature at the outside windows furface when the heat lost per second doubles is  -13.18°C

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Two bar magnets are labeled A and B. The ends of each magnet are numbered 1 or 2, but the poles are not labeled. When A1 is brou
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It is definitely letter D. <span>A1 and B1 are like poles, but there is not enough information to tell whether they are north poles or south poles.

A1 and B1 is either both north poles or both south poles. Repulsion of both magnets says it all--like poles always repel while opposite poles always attract. Thus, the best conclusion to this would be choice D.</span>
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2 years ago
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To show that displacement current is necessary to make Ampère's law consistent for a charging capacitor Ampère's law relates the
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