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Stells [14]
1 year ago
11

The linear expansion coefficient of aluminium is 24 × 10^-6 (°C)^-1. When thetemperature is 33 °C, a spherical aluminium ball ha

s a diameter of 11.21 cm.A) Calculate the diameter (in cm) of the aluminium ball when the temperature is raisedto 257° C? (Round your answer to 3 decimal places.)B) Calculate the change in temperature (in °C) needed to increase the volume of thealuminium ball by 3.24%.
Physics
1 answer:
KatRina [158]1 year ago
6 0

A) Use the following formula for the linear expansion, in this case, of the diameter of the spherical ball:

d=d_o\lbrack1+\alpha(T_2-T_1)\rbrack

where,

do: initial diameter = 11.21cm

d: final diameter = ?

α: linear expansion coefficient = 24*10^6 /°C

T2: final temperature = 257°C

T1: intial temperature = 33°C

Replace the previous values of the parameters into the formula for d and simplify:

Hence, the diameter of the spherical ball is 11.27cm

B) First, consider what is the change in the radius of the sphere, as follow:

\Delta V=0.0324V_o

Now, use the following formula for the change in volume with the temperature:

\Delta V=\beta V_o\Delta T=3\alpha V_o\Delta T

where you have used the equivalence β = 3α.

Solve the equation above for the change in temperature and replace the values of the other parameters:

\Delta T=\frac{\Delta V}{3\alpha V_o}=\frac{0.0324V_o}{3(24\cdot10^{-6}(\degree C)^{-1})V}=450\degree C

Hence, the change in temperature is 450°C

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Answer:

Magnitude the net torque about its axis of rotation is 2.41 Nm

Solution:

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The radius of the wrapped rope around the drum, r = 1.33 m

Force applied to the right side of the drum, F = 4.35 N

The radius of the rope wrapped around the core, r' = 0.51 m

Force on the cylinder in the downward direction, F' = 6.62 N

Now, the magnitude of the net torque is given by:

\tau_{net} = \tau + \tau'

where

\tau = Torque due to Force, F

\tau' = Torque due to Force, F'

tau = F\times r

tau' = F'\times r'

Now,

\tau_{net} = - F\times r + F'\times r'

\tau_{net} = - 4.35\times 1.33 + 6.62\times 0.51 = - 2.41\ Nm

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

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2 years ago
How much heat is removed from 60 grams of steam at 100 °C to change it to 60 grams
Harrizon [31]

Answer:

45200J

Explanation:

Given parameters:

Heat of vaporization of water  = 2260J/g

Mass of steam = 20g

Temperature = 100°C

Unknown:

Energy released during the condensation  = ?

Solution:

This change is a phase change and there is no change in temperature

To find the amount of heat released;

         H  = mL

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Insert the parameters and solve;

         H  = 20g x 2260J/g

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Alex787 [66]

Answer:

The distance between the two objects must be squared.

Explanation:

Gravitational force always act between two objects that have mass. The gravitational force is a weak force and attractive in nature.

The force of pull depends on the masses of the two objects and the distance between them.

The formula to calculate gravitational force between two objects having masses 'm' and 'M' and separated by a distance 'd' is given as:

F_g=\frac{GmM}{d^2}

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Now, from the above formula, it is clear that, the force of gravitation is inversely proportional to the square of the distance between the two objects.

Thus, the quantity that must be squared in the equation of gravitational force between two objects is the distance 'd'.

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It’s C I hope it helps you
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