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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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Gnoma [55]

For this one, all you really need to do is eliminate any answers
that are absurd or meaningless.

You can't increase a transformer.
You can't increase a circuit.
You can't increase a generator.

When the <em><u>current</u></em> through a coil of wire increases, the magnetic field
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7 0
3 years ago
How much will be the net displacement if a car travels 6 km east and then travels 14 km west
ExtremeBDS [4]
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3 years ago
Human reaction times are worsened by alcohol. How much further (in feet) would a drunk driver's car travel before he hits the br
sweet-ann [11.9K]

Answer:

A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes

Explanation:

Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:

2as = Vf² - Vi²

s = (Vf² - Vi²)/2a

where,  

Vf = Final Velocity of Car = 0 mi/h

Vi = Initial Velocity of Car = 50 mi/h

a = deceleration of car  

s = distance covered

Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.

So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:

s = vt

FOR SOBER DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 0.33 s

s = s₁

Therefore,

s₁ = (73.33 ft/s)(0.33 s)

s₁ = 24.2 ft

FOR DRUNK DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 1 s

s = s₂

Therefore,

s₂ = (73.33 ft/s)(1 s)

s₂ = 73.33 ft

Now, the distance traveled by drunk driver's car further than sober driver's car is given by:

ΔS = s₂ - s₁

ΔS = 73.33 ft - 24.2 ft

<u>ΔS = 49.13 ft</u>

6 0
3 years ago
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shtirl [24]

Answer:

Explanation:

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