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masha68 [24]
3 years ago
12

A skier of mass 103 kg comes down a slopeof constant angle 32◦with the horizontal.What is the force on the skier parallel tothe

slope? Neglect friction. The accelerationof gravity is 9.8 m/s2.Answer in units of N.
Physics
1 answer:
LuckyWell [14K]3 years ago
3 0

Answer:

534.9 N

Explanation:

The skier weight is his mass times gravitational acceleration g

W = mg = 103 * 9.8 = 1009.4 N

This weight can be divided into 2 components, one perpendicular and the other parallel to the 32-degree slope. The parallel component would equal to

Wsin(32^0) = 1009.4sin(32^0) = 534.9 N

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An element is a pure substance and is made of only one type of atom; it cannot be broken down into a simpler substance.

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I need help thank you​
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D

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Why are collisions between galaxies more likely than collisions between stars within a galaxy?
KatRina [158]

Answer:

stars share a gravitational force with the galaxy while nearby galaxies do not share a gravitational field.

Explanation:

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If Steve throws the football 50 meters in 3 seconds, what is the average speed of the football? Speed = distance / time Question
leva [86]

Answer:

A. 16.67 m/s

Explanation:

Speed or velocity refers to the rate of change in distance over a change in time. That is;

Speed = Distance ÷ time

Where;

Speed is in metre/seconds

Distance is in metre

Time is in seconds.

In this question, Steve throws a football 50 meters in 3 seconds. The average speed can be calculated this:

S = D/t

Where; d = 50m, t = 3s

S = 50/3

S = 16.6666666

S = 16.67m/s

5 0
3 years ago
A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.6 m/s . Two seconds later t
dybincka [34]

Answer:

A) t = 7.0 s    

B) x = 25 m  

C) v = 10 m/s

Explanation:

The equations for the position and velocity of an object traveling in a straight line is given by the following expressions:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A)When both friends meet, their position is the same:

x bicyclist = x friend

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

If we place the center of the frame of reference at the point when the bicyclist starts following his friend, the initial position of the bicyclist will be 0, and the initial position of the friend will be his position after 2 s:

Position of the friend after 2 s:

x = v · t

x = 3.6 m/s · 2 s = 7.2 m

Then:

1/2 · a · t² = x0 + v · t       v0 of the bicyclist is 0 because he starts from rest.

1/2 · 2.0 m/s² · t² = 7.2 m + 3.6 m/s · t

1  m/s² · t² - 3.6 m/s · t - 7.2 m = 0

Solving the quadratic equation:

t = 5.0 s

It takes the bicyclist (5.0 s + 2.0 s) 7.0 s to catch his friend after he passes him.

B) Using the equation for the position, we can calculate the traveled distance. We can use the equation for the position of the friend, who traveled over 7.0 s.

x = v · t

x = 3.6 m/s · 7.0 s = 25 m

(we would have obtained the same result if we would have used the equation for the position of the bicyclist)

C) Using the equation of velocity:

v = a · t

v = 2.0 m/s² · 5.0 s = 10 m/s

8 0
3 years ago
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