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Anna11 [10]
3 years ago
11

A force of 60 N is applied to a skier to pull him along a horizontal surface so that his speed remains constant. If the coeffici

ent of friction of the skis on snow is 0.05, then what is the weight of the skier?
Physics
1 answer:
matrenka [14]3 years ago
7 0

Answer:

1200\ \text{N}

Explanation:

F = Force on the skier = 60 N

\mu = Coefficient of friction = 0.05

w = Weight of skier

Force is given by

F=\mu w

\Rightarrow w=\dfrac{F}{\mu}

\Rightarrow w=\dfrac{60}{0.05}=\dfrac{6000}{5}

\Rightarrow w=1200\ \text{N}

Weight of the skier on which the force is being applied is 1200\ \text{N} .

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Answer:

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Explanation:

your answer is in this link

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Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has a magnitude of 3.7m/s2.​ We want
lbvjy [14]

Explanation:

It is given that, Onur drops a basketball from a height of 10 m on Mars, where the acceleration due to gravity has a magnitude of 3.7 m/s².

The second equation of kinematics gives the relationship between the height reached and time taken by it.

Here, the ball is droped under the action of gravity. The value of acceleration due to gravity on Mars is positive.

We want to know how many seconds the basketball is in the air before it hits the ground. So, the formula is :

\Delta x=v_ot+\dfrac{1}{2}at^2

t is time taken by the ball to hit the ground

v_o is initial speed of the ball

So, the correct option is (A).

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26. An ice-skater who weighs 200 N is gliding across the ice. If the force of friction is 4 N. what is the
Scrat [10]

Answer:

0.02

Explanation:

coefficient of kinetic friction = μ

force of friction = Ff

Normal Force = FN, but

FN = -W

Ff = -μFN

so μ = Ff/FN

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= 0.02.

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3 years ago
Adam drops a ball from rest from the top floor of a building at the same time Bob throws a ball horizontally from the same locat
guapka [62]

Answer:

Both balls hit the ground at the same time

Explanation:

Adam drops the ball from rest, so the ball just "<em>falls</em>" in vertical direction, being gravity its only acceleration, for cinematic movements we use that:

y(t)=y_{0}+v_{0y}t+\frac{1}{2}gt^{2}

In this case we have that gravity is negative, and as Adam drops the ball, v_{0y}=0

Bob throws the ball horizontally, so the movement will be a <em>parabola</em>, we can divide into horizontal direction, and vertical direction.

But we only need to analize the vertical movement, in wich again the only acceleration is gravity, and compare it with Adam's ball. Again we have that gravity is negative, and as the initial throw is horizontal, v_{0y}=0

Finally, we have that

y(t)=h-\frac{1}{2}gt^{2}

where

h=y_{0}

both for Adam's vertical drop, and for Bob's vertical component of the parabolic throw.

Now, if we put y(t)=0 (the origin of the vertical coordinate), we get for both cases that

h=\frac{1}{2}gt^{2}

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