Answer: N = Mgcos(theta)
Therefore, the Normal reaction force is equal to Mgcos(theta)
Explanation:
See attached for a sketch.
From the attachment.
.
N = normal reaction force on block
W = weight of the block
theta = angle of the inclined plane to the horizontal
From the sketch, we can see that
N is equal in magnitude but opposite direction to Wy
N = Wy
And
Wy = Wcos(theta)
Wx = Wsin(theta)
Then,
N = Wy = Wcos(theta)
But W = mass × acceleration due to gravity = mg
N = Mgcos(theta)
Therefore, the Normal reaction force is equal to Mgcos(theta)
79 m/s. A stone dropped from the top of the Empire State Building will have a velocity of 79 m/s just before it strikes the ground.
This problem is about free fall, to find the velocity of the stone before it strikes the ground we have to use the equation 
, the initial velocity of the stone is 0 m/s. Then:
Solving the equation above with g = 9.8 m/s², and h = 318.0 m:
≅ 79 m/s
Answer:
hope this helps
Explanation:
one in which a positively charged nucleus is surrounded by one or more negatively charged electrons.
Answer:
A 20 N force acting downward on the right end of the lever would keep it horizontal.
Explanation:
Recall that in order to keep the lever horizontal and without rotating, one needs the sum of the acting torques to be zero.
The torque that the 60 N force of the hanging block produces, contributes to a positive torque (according to the convention counter-clock motion is positive), and its magnitude is the product of the force times its distance to the fulcrum: 60 N * 1 m = 60 Nm.
So we need a negative torque of the same absolute magnitude from the right hand side. Then a force downward will accomplish the negative sign required (according to convention clockwise motion is negative). The magnitude of that torque should equal 60 Nm as well, so we find which force, acting at the other end (3 m from the fulcrum) would accomplish such:
Therefore, a 20 N force acting downward would keep the lever horizontal.
