Sally is trying to lift a heavy box, but is not strong enough to lift it. Would the force that she is applying to the box be an
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Answer:
Velocity is 1.73 m/s along 54.65° south of east.
Explanation:
Let unknown velocity be v, mass of billiard ball be m and east direction be positive x axis.
Here momentum is conserved.
Initial momentum = Final momentum
Initial momentum = m x 2i + m x (-1)i = m i
Final momentum = m x v + m x 1.41 j = mv + 1.41 m j
Comparing
mi = mv + 1.41 m j
v = i - 1.41 j
Magnitude of velocity
Direction,
Velocity is 1.73 m/s along 54.65° south of east.
If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.
Reasoning:
It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.
Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.
Thus the electric force applied on the charge in my hand due to each other is,
Here k is the Coulomb constant, and r is the distance between the charges.
It is also known that the force on a positive charge due to another positive charge is acted outwards.
Thus, the force on the charge due to the charge on the east is,
And the force on the charge due to the charge on the north is,
As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of from the north or from the north direction.
Thus the net force is acting in the north-east direction.
Learn more about the direction of the force here,
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