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1) A 5 kg block rests on a flat plane inclined at an angle of 300 to the horizon as shown in the diagram below. What would be the acceleration of the block down the plane assuming the force of friction is negligible?
mgsinQ = ma

2) A projectile launched from the ground landed a horizontal distance of 120.0 m from its launch point. The projectile was in the air for a time of 4.00 seconds. If the projectile landed at the same vertical position from which it was launched, what was the launch speed of the projectile? Ignore air resistance.

3) A 5.0kg solid sphere is in free fall near the surface of the Earth. What is the magnitude of the gravitational force acting on the Earth by the solid sphere? The Earth's mass is 5.98×1024 kg

4) A hypothetical planet orbits a star with mass one-half the mass of our sun. The planet's orbital radius is the same as the Earth's. Approximately how many Earth years does it take for the planet to complete one orbit?

5) During a recent winter storm, bales of hay had to be dropped from an airplane to a herd of cattle below. Assume the airplane flew horizontally at an altitude of 180 m with a constant velocity of 50 m/s and dropped one bale of hay every two seconds. It is reasonable to assume that air resistance will be negligible for this situation. About how far apart from each other will the bales land on the ground?

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Dear student, please refer to the attachment for the solutions. Thank You.

If the frictional force is negligible, and no other external force acts on the block, then the acceleration of the block down the inclined plane is given as
a=g Sinθ=9.8 m.s^(-2)×Sin 30°=4.9m.s^(-2)

Time of flight of a projectile is defined as the time taken by a projectile to return to the same vertical level (at a different horizontal position) from which it is thrown.
Time of flight of a projectile, T= (2v_0 Sinθ)/g [1]
where v0 represents the initial velocity (launch speed) of the projectile.
Range of a projectile is given as, R= (〖v_0〗^2 Sin2θ)/g [2]
In the given problem, time of flight, T = 4sec and range, R = 120m

Therefore, T^2/R= 4^2/120=0.1333
〖((2v_0 Sinθ)/g)〗^2/((〖v_0〗^2 Sin2θ)/g )= (4×〖v_0〗^2×〖Sinθ〗^2×g)/(g^2×〖v_0〗^2×2SinθCosθ)= (2 tanθ)/g = 0.1333
Or, tanθ = 0.1333 x 9.8/2 = 0.6533
Or, θ = tan-1(0.6533) = ...

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