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Bond [772]
2 years ago
7

A river flows at a speed of 3 mph towards sea. A woman in a canoe can paddle through water at a speed of 7 mph. She wishes to pa

ddle for a total time of 2 hours, first upstream (away from the sea) then turning round and paddling back to her starting point. How long should she paddle upstream?
Physics
1 answer:
miss Akunina [59]2 years ago
8 0

Answer:

She can paddle 5.71 miles upstream.

Explanation:

A woman in a canoe can paddle through water at a speed of 7 mph

Speed with which women can paddle = 7 mph

A river flows at a speed of 3 mph towards sea.

Speed of river = 3 mph

She need to paddle for 2 hours.

Let the distance paddled to upstream be d.

We have

    Velocity to upstream = 7 - 3 = 4 mph

    Velocity to downstream = 7 + 3 = 10 mph  

    \texttt{Time for upstream = }\frac{d}{4}\\\\\texttt{Time for downstream = }\frac{d}{10}

We have

          \frac{d}{4}+\frac{d}{10}=2\\\\10d+4d=2\times 4\times 10\\\\14d=80\\\\d=5.71miles

She can paddle 5.71 miles upstream.

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Sophia is planning on going down an 8-m water slide. Her weight is 50 N. She knows that she has gravitational potential energy (
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Answer:

Explanation:

graph would be a straight line from (0, 0) to (400, 8)

Plot points are

PE = mgh

50(0) = 0 J

50(2) = 100 J

50(4) = 200 J

50(6) = 300 J

50(8) = 400 J

4 0
2 years ago
A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the
yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE          

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0     (initial KE = 0 because the ball starts from rest)  (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

3 0
2 years ago
A typical adult human has a mass of about 70 kg.
Misha Larkins [42]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
a. <span>FM GmMmr2
</span>= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2 
<span>= 2.40 x 10-3 N

b. </span><span>FE GmEmr2
= 6.67 x 10-11 N.m2kg 25 .97 x 1034 kg (70kg) 6.38 x 106 m2
=685 N 
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3 0
2 years ago
Denise is conducting a physics experiment to measure the acceleration of a falling object when it slows down and comes to a stop
iren [92.7K]
We need a and we have m and F . Now a = f÷m so therefore a = 4,9 ÷ 0,5 which is 0,98 metres per cubic second
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2 years ago
Read 2 more answers
As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th
NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek

Conservation of Momentum:

P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1

Energy Balance:

\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2}  ....... Eq 2

Substitute Eq 2 into Eq 1

0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V  \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}

Using Eq 1

0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m

7 0
2 years ago
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