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KonstantinChe [14]
3 years ago
12

An electrical drill has 100 J of chemical energy. Of the 100 J, 45 J are transformed into kinetic energy and 20 J are transforme

d into sound. The remaining energy is given off as thermal energy. What is the thermal energy of the drill?
Physics
2 answers:
Dmitry [639]3 years ago
8 0
For the law of conservation of energy, the total output energy must be equal to the total input energy:
E_{in} = E_{out} (1)
The input energy is the chemical energy of the drill:
E_{in}=100 J
The output energy is sum of kinetic energy Ek, sound Es and thermal energy Et:
E_{out} = E_k+E_s+E_t (2)
We know the value of the kinetic energy: E_k =45 J and of the energy converted into sound: E_s=20 J, so we can put together (1) and (2) to find the thermal energy:
E_t=E_{in}-E_k-E_s=100 J-45 J-20 J=35 J
a_sh-v [17]3 years ago
6 0

For the law of conservation of energy, the total output energy must be equal to the total input energy:

(1)

The input energy is the chemical energy of the drill:


The output energy is sum of kinetic energy Ek, sound Es and thermal energy Et:

(2)

We know the value of the kinetic energy:  and of the energy converted into sound: , so we can put together (1) and (2) to find the thermal energy:

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Answer:

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Explanation:

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Since we need to find the height of coffee mug, it's height will not be more than 10 cm.

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Taking the density of air to be 1.29 kg/m3, what is the magnitude of the angular momentum (in kg · m2/s) of a cubic meter of air
Svet_ta [14]

The magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

The rotating equivalent of linear momentum in physics is called angular momentum. Because it is a conserved quantity—the total angular momentum of a closed system stays constant—it is significant in physics. Both the direction and the amplitude of angular momentum are preserved.

Given the density of air to be 1.29 kg/m3 and a wind speed of 73.0 mi/h

We have to find the magnitude of the angular momentum

Let,

ρ = Density of air = 1.29 kg/m^3

v = Speed of wind = 73.0 mi/h = 0.032 km/s

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Let the volume of air be 1 m^3

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Hence the magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

Learn more about angular momentum here:

brainly.com/question/7538238

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