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svet-max [94.6K]
2 years ago
15

A man pushes a crate along a factory floor by exerting a force of 55 N. If

Physics
1 answer:
nika2105 [10]2 years ago
7 0

Answer:

it's 220 J

the second one

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Question 17 options:A 71.8 kg man goes from an area where the acceleration due to gravity is 9.79 m/s2 to an area where the acce
DerKrebs [107]
Mass doesn't change, no matter where you take it.
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3 years ago
A plastic bottle partially filled with water floats on water, even though the density of the plastic (1.2 g/cc) is more than tha
Semmy [17]

Answer:

True the plastic will float because of the principle of flotation or buoyancy

Explanation:

Buoyancy explains it all!!

Buoyancy  is the upward force/upthrust experienced by a body immersed totally or partially in a liquid.

According to the principle of flotation:

<em>"when a body is totally or partially immersed in liquid it experiences an upthrust which is equal to the volume of fluid displaced"</em>

The plastic will float due to the fact the average density of the total volume of the plastic and the air inside it is less than the same volume of water it is floating in

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3 years ago
Where would you find convergent and divergent plate boundaries relative to
Sergio [31]

Answer:

When two tectonic plates meet, we get a “plate boundary.” There are three major types of plate boundaries, each associated with the formation of a variety of geologic features.

Explanation:

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2 years ago
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Dahasolnce [82]

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7 0
3 years ago
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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