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GaryK [48]
2 years ago
12

A large truck is moving at 22.0 m/s. if its momentum is 125,000 kg • meters per second, what is the truck's mass? 176 kg 2750 kg

5680 kg 11,400 kg
Physics
1 answer:
solong [7]2 years ago
7 0

The  mass of the large truck is determined as 5680 kg.

<h3>Mass of the truck</h3>

The mass of the truck is calculated as follows;

P = mv

where;

  • P is momentum
  • m is mass
  • v is velocity

m = P/v

m = 125000/22

m = 5680 kg

Thus, the  mass of the large truck is determined as 5680 kg.

Learn more about momentum here: brainly.com/question/7538238

#SPJ1

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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.

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(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})

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c_{iron}=0.568 J/kg.\°C

c_{iron}=0.6 J/kg.\°C

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