I assume the 100 N force is a pulling force directed up the incline.
The net forces on the block acting parallel and perpendicular to the incline are
∑ F[para] = 100 N - F[friction] = 0
∑ F[perp] = F[normal] - mg cos(30°) = 0
The friction in this case is the maximum static friction - the block is held at rest by static friction, and a minimum 100 N force is required to get the block to start sliding up the incline.
Then
F[friction] = 100 N
F[normal] = mg cos(30°) = (10 kg) (9.8 m/s²) cos(30°) ≈ 84.9 N
If µ is the coefficient of static friction, then
F[friction] = µ F[normal]
⇒ µ = (100 N) / (84.9 N) ≈ 1.2
Explanation:
For equilibrium, 
.
So, 
= 0
= 
= 705.6 N
Also, for equilibrium 
= 0
= 0
or, 
= 
= 176.4 N
Thus, we can conclude that the tension in the first rope is 176.4 N.
Answer:
A) Average speed = 18.75 m/s
B) More time is spent at 15 m/s than at 25 m/s.
Explanation:
Let the first distance be d1 and the second distance be d2.
We are given;
d1 = 10 km = 10000 m
d2 = 10 km = 10000 m
Speed; v1 = 15 m/s
Speed; v2 = 25 m/s
Now, the formula for distance is; Distance = speed x time
Thus:
d1 = v1 x t1
t1 = d1/v1 = 10000/15 = 666.67 seconds
Also,
d2 = v2 x t2
t2 = d2/v2 = 10000/25 = 400 seconds
Average speed = total distance/total time = (10000 + 10000)/(666.67 + 400) = 18.75 m/s
From earlier, since t1 = 666.67 seconds and t2 = 400 seconds, then;
More time at 15 m/s than at 25 m/s.
The force applied to the cannonball and cannon is equal. The explosion inside the cannon will generate a pressure which will turn into a force on both cannonball and cannon. The cannon being heavier and fixed to the ground will move a bit, but the cannonball will be thrown away, fired.
The answer is a property of density. The higher the density, the higher the pressure at the bottom.
Pressure = mass / Area. So given that the 4 samples occupy the same area at the bottom, the mass is going to be the determining factor. Per given volume, mercury has the largest mass. The answer is A
