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expeople1 [14]
3 years ago
13

A solid brass cylinder and a solid wood cylinder have the same radius and mass (the wood cylinder is longer). Released together

from rest, they roll down an incline. (a) Which cylinder reaches the bottom first, or do they tie? (b) The wood cylinder is then shortened to match the length of the brass cylinder, and the brass cylinder is drilled out along its long (central) axis to match the mass of the wood cylinder. Which cylinder now wins the race, or do they tie?
Physics
1 answer:
Lena [83]3 years ago
6 0

Answer:

a. They will be tie

b. Win the wood cylinder

Explanation:

a.

The both cylinders will reach the bottom at the same time notice the relation in the equation in indepent of the length and both have the same radius and the same rotational inertia.

I=\frac{1}{2}*m*r^2

a=\frac{g*sin(\beta)}{1+I_{com}/m*r^2}

So both will be tie

b.

a_{brass}=\frac{g*sin(\beta)}{1+I_{brass}/m*r^2}=a_{wood}=\frac{g*sin(\beta)}{1+I_{wood}/m*r^2}

The acceleration of the wood cylinder is larger than the acceleration of the brass cylinder so the cylinder of wood will reach the bottom first

a_{brass}

So the wood win the race

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The initial temperature of 150 g of ice is ????20°C. The spe- cific heat capacity of ice is 0.5 cal/g·C° and water’s is 1 cal/g·
soldier1979 [14.2K]

1. 13,500 cal

First of all, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. This is given by

Q_1 = m C_i \Delta T

where

m = 150 g is the mass of the ice

C_i = 0.5 cal/g·C° is the specific heat capacity of the ice

\Delta T=0 C-(-20 C)=20^{\circ}C is the change in temperature of the ice

Substituting,

Q_1 = (150 g)(0.5 cal/gC)(20 C)=1500 cal

Now we have to find the amount of heat needed to melt the ice, which is

Q_2 = m \lambda_f

where

m = 150 g is the mass of the ice

\lambda_f = 80 cal/g is the latent heat of fusion

Substituting,

Q_2 = (150 g)(80 cal/g)=12,000 cal

So the total heat required is

Q_3 = 1500 cal + 12,000 cal = 13,500 cal

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The additional amount of heat required to heat the water to 25°C is

Q_4 = m C_w \Delta T

where

m = 150 g is the mass of water

C_w = 1 cal/g·C is the speficic heat capacity of water

\Delta T=25 C-0 C=25^{\circ}C is the change in temperature

Substituting,

Q_4 = (150 g)(1 cal/gC)(25 C)=3,750 cal

3. 9200 cal

First of all, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. As at point 1., this is given by

Q_1 = m C_i \Delta T

where

m = 80 g is the mass of the ice

C_i = 0.5 cal/g·C° is the specific heat capacity of the ice

\Delta T=0 C-(-20 C)=20^{\circ}C is the change in temperature of the ice

Substituting,

Q_1 = (80 g)(0.5 cal/gC)(20 C)=800 cal

Now we have to find the amount of heat needed to melt the ice:

Q_2 = m \lambda_f

where

m = 80 g is the mass of the ice

\lambda_f = 80 cal/g is the latent heat of fusion

Substituting,

Q_2 = (80 g)(80 cal/g)=6,400 cal

Finally, the amount of heat required to heat the water to 25°C is

Q_3 = m C_w \Delta T

where

m = 80 g is the mass of water

C_w = 1 cal/g·C is the speficic heat capacity of water

\Delta T=25 C-0 C=25^{\circ}C is the change in temperature

Substituting,

Q_3 = (80 g)(1 cal/gC)(25 C)=2,000 cal

So the total heat required is

Q=Q_1+Q_2+Q_3=800 cal+6,400 cal+2,000 cal=9,200 cal

4. No

Explanation:

The total heat required for this process consists of 3 different amounts of heat:

1- The heat required to bring the ice at melting temperature

2- The heat required to melt the ice, while its temperature stays constant

3- The heat required to raise the temperature of the water

However, computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 80 g of water by 45°C is not equivalent: in fact, the calculation of point 1) requires to use the specific heat capacity of ice, not that of water, therefore the two are not equivalent.

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