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almond37 [142]
3 years ago
8

James Joule (after whom the unit of energy is named) claimed that the water at the bottom of Niagara Falls should be warmer than

the water at the top, 51 m above the bottom. He reasoned that the falling water would transform its gravitational potential energy at the top into thermal energy at the bottom, where turbulence brings the water almost to a halt.. If this transformation is the only process occurring, how much warmer will the water at the bottom be? The specific heat of water is 4190 J/(kg?K).
Physics
1 answer:
Molodets [167]3 years ago
7 0

Answer:

0.12 K

Explanation:

height, h = 51 m

let the mass of water is m.

Specific heat of water, c = 4190 J/kg K

According to the transformation of energy

Potential energy of water = thermal energy of water

m x g x h = m x c x ΔT

Where, ΔT is the rise in temperature

g x h =  c x ΔT

9.8 x 51 = 4190 x ΔT

ΔT = 0.12 K

Thus, the rise in temperature is 0.12 K.

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Problem 4: a long wire carries current towards east. a positive charge moves westward and just north from the wire. what is the
Alex787 [66]
The direction of the force experienced by the positive charge is upward.

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(if it was a negative charge, we should have taken the opposite direction)
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3 years ago
Which of the following changes would make a heat engine waste more energy as heat
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8 0
3 years ago
A planet's moon travels in an approximately circular orbit of radius 7.0 ✕ 10⁷ m with a period of 6 h 38 min. Calculate the mass
natka813 [3]

Answer:

3.56×10²⁶ Kg.

Explanation:

Note: The gravitational force is acting as the centripetal force.

Fg = Fc........................... Equation 1

Where Fg = gravitational Force, Fc = centripetal force.

Recall,

Fg = GMm/r²......................... Equation 2

Fc = mv²/r............................. Equation 3

Where M = mass of the planet, m = mass of the moon, r = radius of the orbit and G = Universal gravitational constant.

Substituting equation 2 and 3 into equation 1

GMm/r² = mv²/r

Simplifying the equation above,

M = v²r/G .............................. Equation 4.

The period of the moon in the orbit

T = 2πr/v

Making v the subject of the equation,

v = 2πr/T............................. Equation 5

where r = 7.0×10⁷ m, T = 6 h 38 min = (6×3600 + 38×60) s = (21600+2280) s

T = 23880 s, π = 3.14

v = (2×3.14×7.0×10⁷ )/23880

v = 18409 m/s

Also Given: G = 6.67×10⁻¹¹ Nm²/kg²

Also substituting into equation 4

M = 18409²×7.0×10⁷ /(6.67×10⁻¹¹)

M = 3.56×10²⁶ Kg.

Thus the mass of the planet =  3.56×10²⁶ Kg.

5 0
3 years ago
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