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romanna [79]
4 years ago
10

A boy who exerts a 300-N force on the ice of a skating rink is pulled by his friend with a force of 75 N, causing the boy to acc

elerate across the ice. If drag and the friction from the ice apply a force of 5 N on the boy, what is the magnitude of the net force acting on him?
Physics
1 answer:
Bezzdna [24]4 years ago
7 0

Answer:

70 N

Explanation:

Draw a free body diagram of the boy.  There are four forces:

Weight force mg pulling down,

A 300 N normal force pushing up,

A 75 N applied force pulling right,

and a 5 N friction force pushing left.

The boy's acceleration in the y direction is 0, so the net force in the y direction is 0.

The net force in the x direction is 75 N − 5 N = 70 N.

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If an object has a high density then the molicules making up that object are closly packed togeather. Because of this, objects with a higher density will have more mass than objects of the same size that have a lesser density.

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3 years ago
The volume of the water in the graduated cylinder rose as some of the water was displaced by the table tennis ball. Find the vol
Leokris [45]

The approximate volume of table tennis ball is  80 cm³

<h3>What is volume?</h3>

Volume is defined as the amount of space occupied by the three dimensional object. S I unit of volume is m³ or cm³.

To find the volume of tennis ball using graduated cylinder.

Step 1 - Fill the graduated cylinder half or full.

Step 2 - Mark the initial volume of the water i.e. 100 cm³ (Vi)

Step 3  - Put the tennis ball in the graduated cylinder. Some of the water was displaced by the table tennis ball.

Step 4 - Mark the Final volume of the water (Vf) i.e. 180 cm³

Step 5 = Calculate the volume by using Formula

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Hence the volume of tennis ball (Vb) is 80 cm³

For more Volume related question visit here:

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6 0
2 years ago
The closest distance a book can be read from a pair of reading eyeglasses (Power = 1.55 dp) is 26.0 cm. What is the near distanc
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Answer:

The image distance is 20.0 cm.

Explanation:

Given that,

Power = 1.55 dp

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f = \dfrac{1}{P}

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f=\dfrac{1}{1.55}

f=0.645\ m

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Using lens formula

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{-u}

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\dfrac{1}{v}=\dfrac{1}{64.5}-\dfrac{1}{-29}

\dfrac{1}{v}=\dfrac{187}{3741}

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3 years ago
Superman is flying 54.5 m/s when he sees
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Superman's initial velocity is v_{0}=54.5\frac{m}{s}, and he will have to cover a distance d = 850m in a time t = 4.22s. Since we know d, v_{0} and t, we have to find the aceleration a in order to find v_{f}.

From the equation d=v_{0} t+\frac{at^{2} }{2} we have to clear a, getting the equation as follows: a=\frac{2(d-v_{0}t) }{t^{2} }.

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Substituting the values:

v_{f} =54.5\frac{m}{s} +(69.63\frac{m}{s^{2}}.4.22s)=348.34\frac{m}{s}

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