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amm1812
2 years ago
4

 Water flows over a waterfall that is 20 m high at the rate of 4.0 × 104 kg/s. a. How much KE does the water gain each second wh

en it reaches the bottom of the waterfall? b. If this water powers an electric generator with a 40% efficiency, how many watts of electric power can be supplied? C, What is the velocity of the water when it exits the turbine that drives the generator?
Physics
1 answer:
I am Lyosha [343]2 years ago
5 0

Answer:

a)

7.8 x 10⁶ Watt

b)

3.12 x 10⁶ Watt

c)

15.3 m/s

Explanation:

a)

h = height of the waterfall = 20 m

m = mass rate = 40000 kg/s

K = gain in kinetic energy per second

Using conservation of energy

K = mgh

K = (40000) (9.8) (20)

K = 7.8 x 10⁶ J/s

K = 7.8 x 10⁶ Watt

b)

P = Electric power supplied

η = Efficiency = 40% = 0.40

Electric power supplied is given as

P = η K

P = (0.40) (7.8 x 10⁶)

P = 3.12 x 10⁶ Watt

c)

P' = Kinetic energy remaining in water after exiting the turbine

v = velocity of water

Kinetic energy remaining in water after exiting the turbine is given as

P' = K - P

P' = 7.8 x 10⁶ - 3.12 x 10⁶

P' = 4.68 x 10⁶ Watt

(0.5) m v² = 4.68 x 10⁶

(0.5) (40000) v² = 4.68 x 10⁶

v = 15.3 m/s

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11111nata11111 [884]

The spoon to transfer 40 J of energy to your hand is descibed as follows

<u>Explanation:</u>

Given  area of cross section of copper spoon is A = 20mm into 1.5 mm

temperature difference is DT = (100 minus 35) = 65 0C

length of the spoon is l = 18 cm,

amount of heat should be transfer Q = 40 J

coefficient of thermal conductivity of copper k = 400 W by mk

we know that the thermal conductivity is Q by t = k into A into DT by l

t = Q into l by k into A into DT

t = (40 into 0.18) by  \left(400 \times 30 \times 10^{-6} \times 65\right)

t = 9.23 s

6 0
2 years ago
Explain how the forces need to change so the aeroplane can land
Fofino [41]
When an airplane is flying straight and level at a constant speed, the lift it produces balances its weight, and the thrust it produces balances its drag. However, this balance of forces changes as the airplane rises and descends, as it speeds up and slows down, and as it turns.
7 0
2 years ago
You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which was stopped a
bekas [8.4K]

Answer:

A) 12.08 m/s

B) 19.39 m/s

Explanation:

A) Down the hill, we will apply Newton’s second law of motion in the downward direction to get:

mg(sinθ) – F_k = ma

Where; F_k is frictional force due to kinetic friction given by the formula;

F_k = (μ_k) × F_n

F_n is normal force given by mgcosθ

Thus;

F_k = μ_k(mg cosθ)

We now have;

mg(sinθ) – μ_k(mg cosθ) = ma

Dividing through by m to get;

g(sinθ) – μ_k(g cosθ) = a

a = 9.8(sin 12.03) - 0.6(9.8 × cos 12.03)

a = -3.71 m/s²

We are told that distance d = 24.0 m and v_o = 18 m/s

Using newton's 3rd equation of motion, we have;

v = √(v_o² + 2ad)

v = √(18² + (2 × -3.71 × 24))

v = 12.08 m/s

B) Now, μ_k = 0.10

Thus;

a = 9.8(sin 12.03) - 0.1(9.8 × cos 12.03)

a = 1.08 m/s²

Using newton's 3rd equation of motion, we have;

v = √(v_o + 2ad)

v = √(18² + (2 × 1.08 × 24))

v = 19.39 m/s

6 0
2 years ago
A cylinder has 500 cm3 of water. After a mass of 100 grams of sand is poured into the cylinder and all air bubbles are removed b
Nina [5.8K]

Answer: SG = 2.67

Specific gravity of the sand is 2.67

Explanation:

Specific gravity = density of material/density of water

Given;

Mass of sand m = 100g

Volume of sand = volume of water displaced

Vs = 537.5cm^3 - 500 cm^3

Vs = 37.5cm^3

Density of sand = m/Vs = 100g/37.5 cm^3

Ds = 2.67g/cm^3

Density of water Dw = 1.00 g/cm^3

Therefore, the specific gravity of sand is

SG = Ds/Dw

SG = (2.67g/cm^3)/(1.00g/cm^3)

SG = 2.67

Specific gravity of the sand is 2.67

3 0
2 years ago
Stu wanted to calculate the resistance of a light bulb connected to a 4.0-V battery, with a resulting current of 0.5 A. He used
Margarita [4]

Answer:

The answer is not correct.

Explanation:

Stu's answer is not correct, the equation to use is known as the law of ohm. In which the voltage is defined as the product of the current by the resistance, then we will see this equation.

V = I*R\\where:\\I = current [amp]\\R = resistance [ohm]\\V = voltage [volts]\\

In order to find resistance, this term is found multiplying the current on the right side of the equation, therefore the current will be divided on the left side of the equation.

R=\frac{V}{I} \\replacing:\\R=\frac{4}{0.5} \\R=8[ohms]

That is the reason that the result found by Stu is not correct.

6 0
2 years ago
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