Answer:
We can think that the girl is initial over the skateboard, when she jumps forward, most of the force that she applies is downwards, so the force is transmitted from the skateboard to the ground.
Now, as she jumps forward, there is also some force applied in the horizontal direction, this force will affect the skateboard, accelerating it in the opposite direction (backward).
As the skateboard is way less heavy than the girl, the acceleration that the skateboard experiences will be bigger. Now, we can not estimate how much the skateboard moves, because we do not know the initial velocity of the skateboard.
Answer:
See explanation below
Explanation:
In this case, you want to know if you put an object between these forces, which direction would go.
To know this, we need to calculate the moment of an object, which is defined as the product of a force and it's distance. In other words:
M = F * d (1)
And, in order to reach equilibrium the force will exert a direction in clockwise or anticlosewise, and these moments, should be even:
anticlockwise moment = clockwise moment.
The clockwise would be the forces to the right, and anticlock would the only force to the left of the axle.
Clockwise moment = (10 * 0.8) + (25 * 2.6) = 73 Ns
Anticlockwise moment = 34 * 3.5 = 119 Ns.
As we can see, the moment in the anticlockwise is higher than the actual clockwise moment, therefore, we can assume that the object will move anticlockwise, or simply move to the left.
Hope this helps
The correct answer is 1.2 m/s
: mv+mv=mv+mv
(0.5kg)(2m/s)+(0.4kg)(0m/s)=(0.5kg)v+(0.4kg)(1m/s)
= 1kg*m/s=(0.5kg)v+0.4kg*m/s
=1kg*m/s-0.4kg*m/s=(0.5kg)v
=0.6kg*m/s=(0.5kg)v
to solve for v we divide both side by 0.5kg
v=1.2m/s
The height of the ball above the ground is 38.45 m
First we will calculate the velocity of the ball when it touch the ground by using first equation of motion
v=u+gt
v=0+9.81×2.8
v=27.468 m/s
now the height of the ground can be calculated by the formula
v=√2gh
27.468=√2×9.81×h
h=38.45 m
