Static friction keeps the car from skidding off the road and points toward the center of the curve. By Newton's second law, the car experiences
• net vertical force
F [normal] - F [weight] = 0
• net horizontal force
F [friction] = ma = mv²/r
where v is the tangential speed of the car.
It follows that
F [normal] = F [weight] = mg
and when static friction is maximized at the car's maximum speed,
F [friction] = µ F[normal] = 0.402 mg
Solve for v :
0.402 mg = mv²/r ⇒ v = √(0.402 g (93.5 m)) ≈ 19.2 m/s
Heat will be transferred from iron to water, because heat flows always from the higher temperature system to the lower temperature system.
The power that heat pump draws when running will be 6.55 kj/kg
A heat pump is a device that uses the refrigeration cycle to transfer thermal energy from the outside to heat a building (or a portion of a structure).
Given a heat pump used to heat a house runs about one-third of the time. The house is losing heat at an average rate of 22,000 kJ/h and if the COP of the heat pump is 2.8
We have to determine the power the heat pump draws when running.
To solve this question we have to assume that the heat pump is at steady state
Let,
Q₁ = 22000 kj/kg
COP = 2.8
Since heat pump used to heat a house runs about one-third of the time.
So,
Q₁ = 3(22000) = 66000 kj/kg
We known the formula for cop of heat pump which is as follow:
COP = Q₁/ω
2.8 = 66000 / ω
ω = 66000 / 2.8
ω = 6.66 kj/kg
Hence the power that heat pump draws when running will be 6.55 kj/kg
Learn more about heat pump here :
brainly.com/question/1042914
#SPJ4
Answer:
A) 100°C
B) 211 g
Explanation:
Heat released by red hot iron to cool to 100°C = 130 x .45 x 645 [ specific heat of iron is .45 J /g/K]
= 37732.5 J
heat required by water to heat up to 100 °C = 85 x 4.2 x 80 = 28560 J
As this heat is less than the heat supplied by iron so equilibrium temperature will be 100 ° C. Let m g of water is vaporized in the process . Heat required for vaporization = m x 540x4.2 = 2268m J
Heat required to warm the water of 85 g to 100 °C = 85X4.2 X 80 = 28560 J
heat lost = heat gained
37732.5 = 28560 + 2268m
m = 4 g.
So 4 g of water will be vaporized and remaining 81 g of water and 130 g of iron that is total of 211 g will be in the cup . final temp of water will be 100 °C.
