Answer:
60N
Explanation:
in this case the minimum amount of force required must be equal to the friction Force. i.e <u>Newton</u><u>'s</u><u> </u><u>first</u><u> </u><u>law</u><u> of</u><u> </u><u>mot</u><u>ion</u><u>.</u>
therefore the maximum amount of frictional force is equal to the applied force which is 60N.
because of the net force acting on the object is zero the object is in constant motion . i.e equal and opposite force must be applied so that the object is in constant velocity therefore the total frictional force must be 60N
Answer:
balloon pushes you back
Explanation:
3rd Law: Every action has an equal and opposite reaction
So, when you let go of the balloon it's pushed forward so the balloon pushes you back
Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.
Answer:
(A) a net torque but no net force on the loop.
Explanation:
The total force on the loop is zero because the forces on the opposite sides of the loop are equal but act in opposite directions and as a result they cancel each other out. The two forces on opposite sides to the axis of rotation each give rise to a torque about the axis of rotation. This torque is directed along the axis of rotation.
