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juin [17]
3 years ago
9

Do number 9 Please show work

Physics
1 answer:
Mrrafil [7]3 years ago
7 0

Answer:

a) the acceleration of the block is 8.8 m/s^2

b)  the time is 0.33 s

Explanation:

The first that we shall do is to draw the free body diagram, like the one that appears in the attached image. Let's see that the axis x and y were rotated for a better understanding of the equations developed.

We know that the movement with acceleration is the x-axis therefore the total amount of forces will be equal to the product of the mass and the acceleration.

The following step is to calculate the normal force, using the total forces in the y-axis that it is equal to 0. (because there is no movement in the y-axis).

By the definition of the friction force = normal force * friction coefficient. We can calculate the friction force and use it in the equation 1.

For the calculation of the time, we use the kinematics formulas.

Have in count that the body starts moving from the rest therefore v0 = 0

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Question 17 A sample of iron is put into a calorimeter (see sketch at right) that contains of water. The iron sample starts off
ivanzaharov [21]

Complete Question:

A 59.1 g sample of iron is put into a calorimeter (see sketch attached) that contains 100.0 g of water. The iron sample starts off at 85.0 °C and the temperature of the water starts off at 23.0 °C. When the temperature of the water stops changing it's 27.6 °C. The pressure remains constant at 1 atm.

Calculate the specific heat capacity of iron according to this experiment. Be sure your answer is rounded to the correct number of significant digits

(Question attached)

Answer:

c_{iron}=0.568 J/kg.\°C

c_{iron}=0.6 J/kg.\°C (rounded to 1 decimal place)

Explanation:

A calorimeter is used to measure the heat of chemical or physical reactions. The example given in the question is using the calorimeter to determine the specific heat capacity of iron.

When the system reaches equilibrium the iron and water will be the same temperature, T_{e}. The energy lost from the iron will be equal to the energy gained by the water. It is assumed that the only heat exchange is between the iron and water and no exchange with the surroundings.

Q=mc(T_{e}-T_{initial}) (Eq 1)

-Q_{iron}=Q_{water} (Eq 2)

Water:

m_{water}=100.0 g, c_{water}=4.186 J/kg.\°C, T_{initial,water}=23 \°C, T_{e}=27.6 \°C

Iron:

m_{iron}=59.1 g, c_{iron} = ? J/kg.\°C, T_{initial,iron}=85 \°C, T_{e}=27.6 \°C

Substituting Eq 1 into Eq 2 and details extracted from the question:

-m_{iron}c_{iron}(T_{iron,e}-T_{initial})=m_{water}c_{water}(T_{water,e}-T_{initial})

-59.1*c_{iron}(27.6-85)=100.0*4.186(27.6-23)

c_{iron}=0.568 J/kg.\°C

c_{iron}=0.6 J/kg.\°C

4 0
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Answer:

b

Explanation:

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