Answer:
The surface of Mercury has landforms that indicate its crust may have contracted. They are long, sinuous cliffs called lobate scarps. These scarps appear to be the surface expression of thrust faults, where the crust is broken along an inclined plane and pushed upward.
Explanation:
I hope this helps a little bit.
Explanation:
Assuming the wall is frictionless, there are four forces acting on the ladder.
Weight pulling down at the center of the ladder (mg).
Reaction force pushing to the left at the wall (Rw).
Reaction force pushing up at the foot of the ladder (Rf).
Friction force pushing to the right at the foot of the ladder (Ff).
(a) Calculate the reaction force at the wall.
Take the sum of the moments about the foot of the ladder.
∑τ = Iα
Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0
Rw (3.0 sin 60°) = mg (1.5 cos 60°)
Rw = mg / (2 tan 60°)
Rw = (10 kg) (9.8 m/s²) / (2√3)
Rw = 28 N
(b) State the friction at the foot of the ladder.
Take the sum of the forces in the x direction.
∑F = ma
Ff − Rw = 0
Ff = Rw
Ff = 28 N
(c) State the reaction at the foot of the ladder.
Take the sum of the forces in the y direction.
∑F = ma
Rf − mg = 0
Rf = mg
Rf = 98 N
Answer:
The magnitude of the torque is 263.5 N.
Explanation:
Given that,
Applied force = 31 N
Distance from the axis = 8.5 m
She applies her force perpendicularly to a line drawn from the axis of rotation
So, The angle is 90°
We need to calculate the torque
Using formula of torque

Where, F = force
d = distance
Put the value into the formula


Hence, The magnitude of the torque is 263.5 N.
Answer:

Explanation:
v = Velocidad final = 
u = Velocidad inicial = 0
t = Tiempo empleado = 15 s
a = Aceleración
De las ecuaciones cinemáticas tenemos

La aceleración del camión en el primer intervalo de tiempo es
.