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docker41 [41]
3 years ago
7

A frog leaps vertically into the air and encounters no appreciable air resistance.

Physics
1 answer:
r-ruslan [8.4K]3 years ago
6 0
I am pretty sure that the correct answer is C. On the way up, on the way down, and at the highest point its acceleration is 9.8downward. I consider this one to be the right one because <span> the frog is always goes downward due to </span><span> constantly accelerating of gravity. If frog jumps, the upward velocity will be reduced because of force of gravity. Hope everything is clear!</span>
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Hey what language is this?
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3 years ago
what are the variables in an experiment that tests the distance honey flows at different temperatures? identify the independent
soldier1979 [14.2K]

Answer:

The distance that the honey flowed would be the dependent or outcome variable and the temperature of the honey would be the independent variable.

The dependent variable is what is being measured in an experiment. You can remember it by thinking “it depends on what you’re changing.”

The independent variable in an experiment is what is being changed. You can remember this by thinking “the Independent variable is what I as the scientist change.”

Explanation:

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6 0
3 years ago
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12. your friend with great excitement tells you about his newest idea to solve the energy crisis: he wants to use an electromoto
mixas84 [53]
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You can never get more energy out of the electromotor than you put into it,
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Pour yourself a cold glass of soda, then look up "Perpetual Motion" or
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4 0
3 years ago
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stiks02 [169]

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A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long
Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

W=-\mu mg\times l

Put the value into the formula

W=-0.30\times62\times9.8\times3.50

W=-637.98\ J

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}

\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}

Final potential energy is zero

So, \dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl

Put the value into the formula

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50

v_{2}=\sqrt{2\times55.915}

v_{2}=10.57\ m/s

The speed of skier is 10.57 m/s.

Hence,  (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0
3 years ago
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