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Leona [35]
3 years ago
5

small plastic container, called the coolant reservoir, catches the radiator fluid that overflowswhen the automobile engine becom

es hot. The radiator is made of copper, and the coolant has a coefficient of volume expansion of 410 x 10-6 [1/ oC]. If the radiator is filled to its 15 [quart] capacity when the engine is cold at 6.0 [oC], how much overflow from the radiator will spill into the reservoir when the coolant reaches its operating temperature of 92 [oC]
Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
4 0

Answer:

0.53 quart

Explanation:

The volume expansion of the coolant is gotten from ΔV = VγΔθ where ΔV   = change in volume of the coolant, V = initial volume of coolant = 15 quart, γ = coefficient of volume expansion of coolant = 410 × 10⁻⁶ /°C and Δθ = temperature change = θ₂ - θ₁ where θ₁ = initial temperature of coolant = 6 °C and θ₂ = final temperature of coolant = 92 °C. So, Δθ = θ₂ - θ₁ = 92 °C - 6 °C = 86 °C

Since, ΔV = VγΔθ

substituting the values of the variables into the equation, we have

ΔV = VγΔθ

ΔV = 15 × 410 × 10⁻⁶ /°C × 86 °C

ΔV = 528900 × 10⁻⁶ quart

ΔV = 0.528900 quart

ΔV ≅ 0.53 quart

Since the change in volume of the coolant equals the spill over volume, thus the overflow from the radiator will spill into the reservoir when the coolant reaches its operating temperature of 92 °C is 0.53 quart.

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Answer

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impulse = change in momentum

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             = -12.6 Kg m/s

Magnitude of impulse = 12.6 Kg m/s

b)                                                        

Force = \dfrac{impulse}{time}

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Force = 2520 N

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Some of the highest tides in the world occur in the Bay of Fundy on the Atlantic Coast of Canada. At Hopewell Cape the water dep
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Explanation:

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Therefore, to find the velocity of the depth displacement with time, we differentiate the given expression with respect to time as follows;

D'(t) = \frac{d(7 + 5\cdot cos[0.503(t-6.75)])}{dt}

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= -2.515×(-sin(0.503×t-3.395))

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(a) at 5:00 AM = 5 -  0:00 = 5

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(b) at 6:00 AM = 6 -  0:00 = 6

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(d) at Noon 12:00 PM = 12 -  0:00 = 12

D'(5) =  -2.515×(-sin(0.503×12-3.395)) = -1.21 m/h.

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