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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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Maurinko [17]

This question is not complete.

The complete question is as follows:

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?

Explanation:

a. Using the expression;

T = 2π√R/g

where R = radius of the space = diameter/2

R = 800/2 = 400m

g= acceleration due to gravity = 9.8m/s^2

1/T = number of revolutions per second

T = 2π√R/g

T = 2 x 3.14 x √400/9.8

T = 6.28 x 6.39 = 40.13

1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute

6 0
3 years ago
Steam at 100°C is condensed into a 38.0 g aluminum calorimeter cup containing 280 g of water at 25.0°C. Determine the amount of
KonstantinChe [14]

Answer:

7.2g

Explanation:

From the expression of latent heat of steam, we have

Heat supplied by steam = Heat gain water + Heat gain by calorimeter

mathematically,

m_{s}c_{w} \alpha _{w} + m_{s}l=m_{w}c_{w} \alpha _{w} +m_{c}c_{c} \alpha c_{c}

L=specific latent heat of water(steam)=2268J/g

c_{w}=specific heat capacity=4.2J/gK

c_{c}=specific heat capacity of calorimeter =0.9J/gk

m_{w}=280g

m_{c}=38g

α=change in temperature

\alpha _{c}=(40-25)=15

\alpha _{w}=(40-25)=15

\alpha _{s}=(100-40)=60

Note: the temperature of the calorimeter is the temperature of it content.

From the equation, we can make m_{s} the subject of formula

m_{s}=\frac{m_{w}c_{w} \alpha +m_{c}c_{c}\alpha}{c_{w}\alpha +l}

Hence

m_{s}=\frac{(280*4.2*15) +(38*0.9*15)}{(4.2*60) +2268} \\m_{s}=\frac{18153}{2520}\\ m_{s}=7.2g

Hence the amount of steam needed is 7.2g

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