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3 years ago

95 POINTS PLEASE HELP SIMPLE QUESTION

Physics

1 answer:

gtnhenbr [62]3 years ago

3 0

Given that,

Combined weight of a student and the waxed skis, W = 850 N

It should be given that the coefficient of friction is 0.05.

To find,

The magnitude of the normal force exerted and the magnitude of the force of friction.

Solution,

Normal force = weight of the object,

Here, weight = 850 N

Normal force = 850 N

Let f is the force of friction. It can be calculated as follows :

$f=\mu N\\\\=0.05\times 850\\\\=42.5\ N$

Hence, the normal force is 850 N and the magnitude of the force of friction is 850 N.

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Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe

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Answer:

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Explanation:

Given:

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$y_i$ = initial height of the ball from where Julie throws the ball = 1.5 m
$y_f$ = final position of the ball where Sarah catches the ball = 1.5 m

Let us assume the following:

$u_x$ = horizontal component of the initial speed
$u_y$ = vertical component of the initial speed
$a_x$ = horizontal acceleration of the ball
$a_y$ = vertical acceleration of the ball

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This means if we break the motion of the ball along two axes and study it, we have a uniform acceleration motion in the vertical direction and a zero acceleration motion along the horizontal.

Since the ball has a zero acceleration motion along the horizontal axis, the ball must have a constant speed along the horizontal at all instant of time.

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As this horizontal velocity remains constant in the horizontal motion at all instants of time. So, the horizontal component of the ball's velocity when Sarah catches the ball is 14.12 m/s.

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3 years ago

Read 2 more answers

A rock of mass 0.340 kg is spun horizontally at the end of a wire that has a diameter of 1.00 mm. If the wire gets stretched by

borishaifa [10]

Answer:

Y = 78.13 x 10⁹ Pa = 78.13 GPa

Explanation:

First we will find the centripetal force acting on the wire as follows:

F = mv²/r

where,

F = Force = ?

m = mass of rock = 0.34 kg

v = speed = 19 m/s

r = length of wire

Therefore,

F = (0.34)(19)²/r

F = 122.74/r

now, we find cross-sectional area of wire:

A = πd²/4

where,

A = Area = ?

d = diameter of wire = 1 mm = 0.001 m

Therefore,

A = π(0.001)²/4

A = 7.85 x 10⁻⁷ m²

Now, we calculate the stress on wire:

Stress = F/A

Stress = (122.74/r)/(7.85 x 10⁻⁷)

Stress = 1.56 x 10⁸/r

Now, we calculate strain:

Strain = Δr/r

where,

Δr = stretch in length = 2 mm = 0.002 m

Therefore,

Strain = 0.002/r

now, for Young's modulus (Y):

Y = Stress/Strain

Y = (1.56 x 10⁸/r)/(0.002/r)

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