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stiv31 [10]
3 years ago
15

an object of mass 4 kg is placed on an inclined plane at an angle of 60. if the object slides down the plane with an acceleratio

n of 3m/s, what is the coefficient of the kinetic friction
Physics
1 answer:
balandron [24]3 years ago
6 0

The forces on the  y axis are:

N-mgcos(60)=0   , wich becomes

N=mgcos(60)

Rember that the friction force is always contrary to the motion of an object and its formula is  f=μ * N

The forces in the x axis are:

-f + mgsin(60)= m * a

-μ*mgcos(60) + mgsin(60)=m*a  ,

μ = ( m*a - mgf=μ[sin(60) )/ ( mgcos(90) )


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Sound travels slowest through gas because the atoms in a gas are further apart than the other states of matter.
Tanzania [10]

Answer:

false

Explanation:

If they were farther apart they would be able to slip by through which means that it can go by faster. If the atoms where closer together then yes they would be able to go by slower.

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2 years ago
7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
wlad13 [49]

Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

3 0
3 years ago
A 75.0-kg person climbs stairs, gaining 2.50 m in height. Find the work done to accomplish this task.
ololo11 [35]

Answer:

W=1837.5J

Explanation:

A force exerts work when there is a displacement of its point of application in the direction of that force. Therefore, the work done by a system is defined as the inner product between the applied force and the displacement:

W=\vec{F}\cdot \vec{d}\\W=Fcos\theta d

In this case, we have:

F=mg\\h=dcos\theta

So, replacing this:

W=mgh\\W=75kg*9.8\frac{m}{s^2}*2.5m\\W=1837.5J

4 0
3 years ago
If a car is moving to the left with constantvelocity, one can conclude that
jeka94

Answer:

The net force applied to the car is zero.

Explanation:

We are given that a car is moving to the left with constant velocity.

When the car moving with constant velocity

Then, the final velocity=Initial velocity

Change in velocity=Final velocity- initial velocity=0

When change in velocity is zero then , acceleration of car

a=\frac{change\;in\;velocity}{time}=\frac{0}{t}=0

When acceleration is zero then, By Newtons second law

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The net force applied on the car will be zero.

Option C:The net force applied to the car is zero.

5 0
3 years ago
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