If the pulling is done parallel to the floor with constant velocity, then the box is in equilibrium. In particular, the weight and normal force cancel, so that
<em>n</em> = 38 N
The friction force is proportional to the normal force by a factor of 0.27, so that
<em>f</em> = 0.27 (38 N) ≈ 10.3 N
and so the answer is D.
I got you b, V(final)^2=V(initial+2acceleration*displacement
So this turns to (0m/s)^2=(50m/s)^2+2(9.8)(d) so just flip it all around to isolate d so you get
-(50m/s)^2/2(9.8) = d so you get roughly 12.7555 meters up
Answer:
20 cm
Explanation:
Given that a ball is released from a vertical height of 20 cm. It rolls down a "perfectly frictionless" ramp and up a similar ramp. What vertical height on the second ramp will the ball reach before it starts to roll back down?
Since it is perfectly frictionless, the Kinetic energy in which the ball is rolling will be equal to the potential energy at the edge of the ramp.
Therefore, the ball will reach 20 cm before it starts to roll back down.
Answer:
θ=142.9°
Explanation:
d=1 *r
angle ϕ= 37.1°
the line connecting pebble and target should be tangent to a circle so
cos(180-ϕ-θ)=
=
∴ θ=180-ϕ-
θ= 180-37.1-0
θ=142.9°
D is the answer because your body can do that in all the 7 location where the heat goes in your body in 2,3, OR 5-6 days
