Answer:
0.75
Explanation:
Since the static frictional force is the maximum force applied just before sliding, our frictional force, F is 300 N.
Since F = μN where μ = coefficient of static friction and N = normal force = 400 N (which is the downward force applied against the surface).
So, μ = F/N
= 300 N/400 N
= 3/4
= 0.75
So, the coefficient of static friction μ = 0.75
<u>Explanation:</u>
Velocity of B₁ = 4.3m/s
Velocity of B₂ = -4.3m/s
For perfectly elastic collision:, momentum is conserved
where,
m₁ = mass of Ball 1
m₂ = mass of Ball 2
v₁ = initial velocity of Ball 1
v₂ = initial velocity of ball 2
v'₁ = final velocity of ball 1
v'₂ = final velocity of ball 2
The final velocity of the balls after head on elastic collision would be
Substituting the velocities in the equation
If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.
Answer:
a
Explanation:
heat is energy, energy cannot be made or destroyed but transferred
, to hit nails into a piece of wood or a wall, or to break things into pieces.
Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
