Answers:
B.) 
C.) 
Explanation:
The image attached shows the way the force 
is acting on the block. Now, if we draw a free body diagram of the situation and write the equations for the Net Force in X and Y, we will have the following:
Net Force in X:
(1)
Where:
is the Normal force
is the magnitude of the force exerted on the block
is the angle
Net Force in Y:
(2)
Where:
is the Friction force (it is expresed with the 
sign because this force may be up or down, we cannot know because the block is at rest)
is the gravity force
Rewrittin (1):
(3) This is according to option B
Rewritting (2):
(3) This is according to option C
They were published in 1542.
Answer:
Part a)
Part b)
T = 4.68 s
Explanation:
Part a)
Shell is fired at speed of 40 m/s at angle of 35 degree
so here we have
since gravity act opposite to vertical speed of the shell so at the highest point of its trajectory the vertical component of the speed will become zero
so at the highest point the speed is given
Part b)
After completing the motion we know that the displacement of the object will be zero in Y direction
so we have
The answer is c hope it helps
The answer is 1,600 J.
A work (W) can be expressed as a product of a force (F) and a
distance (d):
W = F · d<span>
We have:
W = ?
F = 20 N = 20 kg*m/s</span>²
d = 80 m
_____
W = 20 kg*m/s² * 80 m
W = 20 * 80 kg*m/s² * m
W = 1600 kg*m²/s²
W = 1600 J
