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NeX [460]
3 years ago
11

Niagara Falls is a set of very large waterfalls located on the border between New York and Ontario, Canada. Over 200,000 cubic f

eet of fast-moving water falls approximately 180 feet every second. Water at the top of the Falls possesses
kinetic energy and gravitational potential energy.
only gravitational potential energy.
only kinetic energy.
neither gravitational potential energy nor kinetic energy.
Physics
2 answers:
Minchanka [31]3 years ago
7 0
The first choice is correct.  It has both.

-- It has kinetic energy because it's "fast-moving" water.

-- It has gravitational potential energy because it's hundreds of feet
above the riverbed that it's getting ready to fall down to, as soon as
it goes over the edge.
vladimir1956 [14]3 years ago
3 0
Yep its A-kinetic energy and gravitational potential energy
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Professional Application: A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 11
Nataliya [291]

Answer:

0.182 m/s

Explanation:

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let the velocity of loaded freight car is v

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An electron is released from rest at a distance of 0.470 m from a large insulating sheet of charge that has uniform surface char
ArbitrLikvidat [17]

Answer:

Part a)

W = 1.58 \times 10^{-20} J

Part b)

v = 1.86 \times 10^5 m/s

Explanation:

Part a)

Electric field due to large sheet is given as

E = \frac{\sigma}{2\epsilon_0}

\sigma = 4.00 \times 10^{-12} C/m^2

now the electric field is given as

E = \frac{4.00 \times 10^{-12}}{2(8.85 \times 10^{-12})}

E = 0.225 N/C

Now acceleration of an electron due to this electric field is given as

a = \frac{eE}{m}

a = \frac{(1.6 \times 10^{-19})(0.225)}{9.1 \times 10^{-31}}

a = 3.97 \times 10^{10}

Now work done on the electron due to this electric field

W = F.d

d = 0.470 - 0.03

d = 0.44 m

So work done is given as

W = (ma)(0.44)

W = (9.11 \times 10^{-31})(3.97 \times 10^{10})(0.44)

W = 1.58 \times 10^{-20} J

Part b)

Now we know that work done by all forces = change in kinetic energy of the electron

so we will have

W = \frac{1}{2}mv^2 - 0

1.58 \times 10^{-20} = \frac{1}{2}(9.1\times 10^{-31})v^2

v = 1.86 \times 10^5 m/s

7 0
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