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Dmitry [639]
3 years ago
11

What is the kinetic energy of a 0.150 kg object moving at a velocity of 100m/s?

Physics
1 answer:
ozzi3 years ago
7 0

Answer:

750J

Explanation:

Given parameters:

Mass of the object   = 0.15kg

Velocity of the object  = 100m/s

Unknown:

Kinetic energy of the body  = ?

Solution:

The kinetic energy is the energy due to the motion of a body. It is mathematically expressed as;

        K.E  = \frac{1}{2} m v²  

m is the mass of the body

v is the velocity

 Now insert the parameters and solve;

     K.E  = \frac{1}{2}  x 0.15 x 100²   = 750J

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The electric current in a wire is 1.5A. How many electrons flow past a given point in a time of 2s?
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Answer:

The amount of electrons that flow in the given time is 3.0 C.

Explanation:

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i.e           I = \frac{Q}{t} ...................(1)

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In the given question, I = 1.5A, t = 2s, find Q.

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                = 1.5 × 2

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3 years ago
In a periodic table a set of properties repeats from?
bogdanovich [222]

Answer;

Row to row

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The heat loss from a boiler is to be held at a maximum of 900Btu/h ft2 of wall area. What thickness of asbestos (k= 0.10 Btu/h f
zmey [24]

Answer:

a. 0.122 ft b. -70 Btu/h ft² c. 633.33 °F

Explanation:

a. Since the rate of heat loss dQ/dt = kAΔT/d where k = thermal conductivity, A = area, ΔT = temperature gradient and d = thickness of insulation.

Now [dQ/dt]/A = kΔT/d

Given that [dQ/dt]/A = rate of heat loss per unit area = -900Btu/h ft², k = 0.10 Btu/h ft ℉(for asbestos), ΔT = T₂ - T₁ = 500 °F - 1600 °F = -1100 °F. We need to find the thickness of asbestos, d. So,

d = kΔT/[dQ/dt]/A

d = 0.10 Btu/h ft ℉ × -1100 °F/-900Btu/h ft²

d = 0.122 ft

b. If the 3 in thick Kaolin is added to the outside of the asbestos, and the outside temperature of the asbestos is 250℉, the heat loss due to the Kaolin is thus

[dQ/dt]/A = k'ΔT'/d'

k' = 0.07 Btu/h ft ℉(for Kaolin), ΔT' = T₂ - T₁ = 250 °F - 500 °F = -250 °F and d' = 3 in = 3/12 ft = 0.25 ft

[dQ/dt]/A = 0.07 Btu/h ft ℉ × -250 °F/0.25 ft

[dQ/dt]/A  = -70 Btu/h ft²

c. To find the temperature at the interface, the total heat flux equals the individual heat loss from the asbestos and kaolin. So

[dQ/dt]/A = k(T₂ - T₁)/d + k'(T₃ - T₂)/d' where  [dQ/dt]/A = -900Btu/h ft², k = 0.10 Btu/h ft ℉(for asbestos), k' = 0.07 Btu/h ft ℉(for Kaolin), T₁ = 1600 °F, T₂ = unknown and T₃ = 250℉.

Substituting these values into the equation, we have

-900Btu/h ft² = 0.10 Btu/h ft ℉(T₂ - 1600 °F)/0.122 ft + 0.07 Btu/h ft ℉(250℉ - T₂)/0.25 ft

-900Btu/h ft² = 0.82 Btu/h ft ℉(T₂ - 1600 °F) + 0.28Btu/h ft ℉(250℉ - T₂)

-900 °F = 0.82(T₂ - 1600 °F) + 0.28(250℉ - T₂)

-900 °F = 0.82T₂  - 1312°F + 70 °F - 0.28T₂

collecting like terms, we have

-900 °F + 1312°F - 70 °F = 0.82T₂   - 0.28T₂

342 °F = 0.54T₂

Dividing both sides by 0.54, we have

T₂ = 342 °F/0.54

T₂ = 633.33 °F

8 0
3 years ago
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