Answer:
C.) A high velocity and Large mass.
Explanation:
Momentum of any object is defined by following formula
Here
: m = mass of object
v = velocity of object
now we know that since momentum is product of mass and velocity
So in order to have more momentum we need the value of this product to be more. So this product will me large is both the physical quantity will be more in magnitude. So if mass is large and velocity will be more then the product of them will be large and hence the momentum of object will be more. Btw I had that question too.
Impulse = change in momentum
The car's momentum was (mass) x (speed)
Momentum = (2400 kg) x (20 m/s)
Momentum = 48,000 km-m/s
To completely stop the car, the impulse = -48,000 km-m/s .
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.
If he keeps that pace he will be at the 34 yard line