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Nimfa-mama [501]
3 years ago
12

A skier is traveling down a slope that is inclined 30° above the horizontal. The coefficient of kinetic friction between the ski

er and the slope is 0.10.
Which of the following best describes the acceleration of the skier?A) It is zero. B) It is about 4.0 m/s^2 C) It is about 9.0 m/s^2 D) It is about 10 m/s^2E) It cannot be determined without knowing the mass of the skier.
Physics
1 answer:
Dennis_Churaev [7]3 years ago
4 0

Answer:

It cannot be determined without knowing the mass of the skier.

Explanation:

A skier is traveling down a slope that is inclined 30° above the horizontal. The coefficient of kinetic friction between the skier and the slope is 0.10.

Which of the following best describes the acceleration of the skier?A) It is zero. B) It is about 4.0 m/s^2 C) It is about 9.0 m/s^2 D) It is about 10 m/s^2E) It cannot be determined without knowing the mass of the skier.

u=f/R

f=frictional force over reaction.masinalpha

U=coefficient of Kinetic friction

R=reaction

f=ma, since the frictional force is in the opposite direction of the force applied to move down the inclined plane

from newtons second law of motion , we know that the force applied is directly proportional with rate of change in momentum.

to get the acceleration , acceleration , mass needs to be given as force is dependent on the mass acceleration.

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Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > 38  \frac{ft}{s^2}  x  \frac{1mi}{5280ft} x  \frac{(3600s)^2}{(1h)^2} = 93272.27  \frac{mi}{h^2}

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > v_2^2=v_1^2 + 2as

The initial velocity is 50mi/h (v_1=50)

When it stops the final velocity is (v_2=0)

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s =  2500/(185644.54)

s=0.0134


So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 0.0134 mi *  \frac{5280ft}{1mi}  = 70.8 ft
So this means that the car traveled in feet 70.8 ft before it came to a stop.

4 0
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A man travles at an average velocity of 12 m/s for 4 seconds. what is the total distance travled by the man?
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Answer:

A. h = 2.15 m

B. Pb' = 122 KPa

Explanation:

The computation is shown below:

a)  Let us assume the depth be h

As we know that

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After solving this,  

h = 2.15 m

Therefore the depth of the fluid is 2.15 m

b)

Given that  

height of the extra fluid is

h' = \frac{2.35 \times 10^{-3}}{ area} \\\\ h' = \frac{2.35 \times 10^{-3}} { 66.2 \times 10^{-4}}

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