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Ksju [112]
4 years ago
11

Increasing the amplitude of a traveling wave with a period T will

Physics
1 answer:
goblinko [34]4 years ago
6 0
<span>Amplitude is the maximum extent of vibration
As by increasing the amplitude the wave speed ,frequency and wavelength will remain same so it will increase </span><span>the speed of the individual particle in the medium. 
This will increase wave energy
so correct option is B
hope it helps</span>
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The electric output of a power plant is 716 MW. Cooling water is the main way heat from the powerplant is rejected, and it flows
Stels [109]

Answer:

(a) 83475 MW

(b) 85.8 %

Explanation:

Output power = 716 MW = 716 x 10^6 W

Amount of water flows, V = 1.35 x 10^8 L = 1.35 x 10^8 x 10^-3 m^3

mass of water, m = Volume  x density = 1.35 x 10^8 x 10^-3 x 1000

                                                               = 1.35 x 10^8 kg

Time, t = 1 hr = 3600 second

T1 = 25.4° C, T2 = 30.7° C

Specific heat of water, c = 4200 J/kg°C

(a) Total energy, Q = m x c x ΔT

Q = 1.35 x 10^8 x 4200 x (30.7 - 25.4) = 3 x 10^12 J

Power = Energy / time

Power input = P = \frac{3 \times 10^{12}}{3600}=8.35 \times 10^{8}W

Power input = 83475 MW

(b) The efficiency of the plant is defined as the ratio of output power to the input power.

\eta =\frac{Power output}{Power input}

\eta =\frac{716}{83475}=0.858

Thus, the efficiency is 85.8 %.

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3 years ago
A batter swings at a baseball. The action force is the bat hitting the ball with a force of 5N. The reaction force, however, can
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This is false. The reaction force can be determined.
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Which statement accurately describes impulse?<br> State corrrect answer from the choices
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Explanation:

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Read 2 more answers
Find the useful power output (in W) of an elevator motor that lifts a 2600 kg load a height of 30.0 m in 12.0 s, if it also incr
Annette [7]

Answer:

P = 251,916.667 W

Cost = 2,267.25 cents

Explanation:

To solve this question we will use the Work Energy Theorem, which is

W = dP + dK\\

Where

dP = Change in Potential Energy

dK = Change in Kinetic Energy

Change in Potential Energy

P_{i} = mgh_{i}\\  P_{f} = mgh_{f}

Where

P_{i} = Initial Potential Energy

P_{f} = Final Potential Energy

m = Mass of System = 10,000 kg

g = Acceleration due to gravity = 9.81 m/s

h_{i} = Initial Height = 0

h_{f} = Final Height = 30 m

Inputting the values we get the answer for dP

dP = P_{f} - P_{i}\\dP= mgh_{f} - mgh_{i}\\ dP= 10000(9.81)(30) - 0\\ dP= 2943000

Change in Kinetic Energy

K_{i} = \frac{1}{2} mv_{i} ^2\\ K_{f} = \frac{1}{2} mv_{f} ^2

Where

K_{i} = Initial Kinetic Energy

K_{f} = Final Kinetic Energy

m = Mass of System = 10,000 kg

g = Acceleration due to gravity = 9.81 m/s

v_{i} = Initial Velocity = 0 m/2

v_{f} = Final Velocity = 4 m/s

Inputting the values we get the answer for dK

dK = K_{f} - K_{i}\\ dK = \frac{1}{2} mv_{f} ^2 - \frac{1}{2} mv_{i} ^2\\ dK = \frac{1}{2} (10000)(4)^2 - 0 \\ dK = 80000

Total Work

W = dP + dK\\

Inputting the values

W = 2943000 + 80000

W = 3,023,000

a) Finding the useful Power Output

P = \frac{W}{t}

Where

P = Power Output

W = Work Done = 3,023,000J

t = Time = 12s

Inputting the values

P = \frac{3,023,000}{12}\\ P = 251,916.667

P = 251,916.667 W

b) Finding the Total Cost

Cost = $0.0900 x P/1000

Cost = $0.0900 x (251,916.667/1000)

Cost = $22.67 or 2,267.25 cents

4 0
4 years ago
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