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Dafna11 [192]
3 years ago
5

A wheel of mass M and radius R rolls on a level surface without slipping. If the angular velocity of the wheel about its center

is ?, what is its linear momentum relative to the surface?
Physics
1 answer:
Verizon [17]3 years ago
4 0
The most important thing you should remember in order to get how is its linear momentum relative to the surface is that the<span> point on the circumference is moving at w rad/second. 
Therefore, here is the solving formulae:
if </span>v = R * w you can easily get <span> linear momentum p :
</span> p = M * v = M * R * w
I'm sure that helps.
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Frictional force and Applied force has same “magnitude” and “opposite” direction.  

Option: B  

<u>Explanation</u>:  

When a book is moved horizontally by applying “force” on the book, the frictional force is opposed to the book by the table. Here, this “frictional force” is opposing the book has the same force what we applied on the book but this frictional force and the applied force are opposite in direction. Always the “frictional force” is opposite to the “applied force” which stops the object to move. For example, if a force applied leftward to the object the frictional force is acted on the right side of the object.

When two objects are in contact they experience a "frictional force". This "frictional force" acts opposite to the force applied on to move the object.

Formula for "frictional force" is \mu\times N

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Which of the following is a scalar? centripetal force<br> weight<br> speed<br> acceleration
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6 0
3 years ago
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Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, t
wolverine [178]

Answer:

final velocity =  0.08585m/s

Explanation:

We are taking train cars as our system. In this system no external force is acting. So we can apply the law of conservation of linear momentum.

The law of conservation of linear momentum states that the total linear momentum of a system remains constant if there is no external force acting on the system. That is total linear momentum before = total linear momentum after

total linear momentum before = linear momentum of first train car + linear momentum of second train car

We know that linear momentum = mv

where,

m = mass

v = velocity

thus,

total linear momentum before = m₁v₁ + m₂v₂

m₁ = mass of first train car = 135,000kg

v₁ = velocity of first train car = 0.305m/s

m₂ = mass of first second car =  100,000kg

v₂ = velocity of second train car =  −0.210m/s

Note: Momentum is a vector. So while adding momentum we should take account of its direction too. Here since second train car is moving in a direction opposite to that of the first one, we have taken its velocity as negative.

total linear momentum before = m₁v₁ + m₂v₂

                                                  = 135,000x0.305 + 100,000x(−0.210)

                                                  = 135,000x0.305 - 100,000x0.210

                                                  = 20,175 kgm/s

Now we have to find total linear momentum after bumping. After the bumping both the train cars will be moving together with a common velocity(say v).

Therefore, total linear momentum after = mv

m = m₁ + m₂ = 135,000 + 100,000 = 235,000

total linear momentum before = total linear momentum after

235,000v = 20,175

v =  \frac{20,175}{235,000}

  = 0.08585m/s

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