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3 years ago

12

Water flows over a section of Niagara Falls at the rate of 1.2×106 kg/s and falls 50.0 m. How much power is generated by the fal

ling water?

1 answer:

quester [9]3 years ago

7 0

Answer:

5.88×10⁸ W

Explanation:

Power = energy / time

P = mgh / t

P = (m/t) gh

P = (1.2×10⁶ kg/s) (9.8 m/s²) (50.0 m)

P = 5.88×10⁸ W

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2 years ago

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Matt and Anna Killian are frequent fliers on Fast-n-Go Airlines. They often fly between two cities that are a distance of 1575

marin [14]

Answer:

Speed of wind = 50mi/hr, Speed of plane in still air = 400mi/hr

Explanation:

Let the speed of the wind = Vw,

Speed of the plane in still air = Vsa,

The first trip the average speed of the plane = 1575mi/4.5hours = 350mi/hr

The coming trip the wind behind = 1575mi/3.5hrs = 450

Write the motion in equation form

First trip ( the plane flew into the wind)

Vaverage = Vsa - Vw

350 = Vsa - Vw

Second trip the wind was behind

450 = Vsa +Vw

Adding the two equation

800 = 2Vas

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Substitute for Vas into equation 1

350mi/hr = 400mi/hr - Vw

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3 years ago

Can u read a ruler it's very hard

iris [78.8K]

Answer:

A. 1 inch

B. 1/2 inch

C. 3/4 inch

D. 1/8 inch

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3 years ago

A horizontal spring with spring constant of 9.80 N/m is attached to a block with a mass of 1.20 kg that sits on a frictionless s

Andreas93 [3]

Answer:

Explanation:

Given

at certain point displacement and velocity is 0.345 m and 0.54 m/s

Therefore Potential and Kinetic Energy associated is

$U=\frac{1}{2}kx^2$

$U=\frac{1}{2}\times 9.8\times (0.345)^2$

$U=0.583 J$

Kinetic Energy $K=\frac{1}{2}mv^2$

$k=\frac{1}{2}\times 1.2\times (0.54)^2=0.174 J$

Total Energy $T=U+K=0.583+0.174=0.757 J$

Total Energy is conserved hence at maximum displacement all energy will be Potential energy

$T=\frac{1}{2}kA^2$

where A=maximum displacement

$0.757=\frac{1}{2}\times 9.8\times A^2$

$0.1544=A^2$

$A=0.393 m$

Maximum speed occurs at equilibrium Position where Potential Energy is zero

thus $T=\frac{1}{2}mv^2$

$0.757=\frac{1}{2}\times 1.2\times v_{max}^2$

$v_{max}=1.123 m/s$

(c)When block is at 0.2 m from Equilibrium speed then its Potential Energy is

$U=\frac{1}{2}kx^2=\frac{1}{2}\times 9.8\times (0.2)^2=0.196 J$

$T=U+K$

$K=0.757-0.196=0.561 J$

$K=\frac{1}{2}mv^2$

$0.561=\frac{1}{2}\times 1.2\times v^2$

$v=0.966 m/s$

7 0

3 years ago