S=Vt
110=V(72)
110/72=V
V=1.527m/s
Given that force is applied at an angle of 30 degree below the horizontal
So let say force applied if F
now its two components are given as
Now the normal force on the block is given as
now the friction force on the cart is given as
now if cart moves with constant speed then net force on cart must be zero
so now we have
so the force must be 199.2 N
Answer:
Magnets come in a variety of shapes and one of the more common is the horseshoe (U) magnet. The horseshoe magnet has north and south poles just like a bar magnet but the magnet is curved so the poles lie in the same plane. The magnetic lines of force flow from pole to pole just like in the bar magnet.
a. The restoring force in the spring has magnitude
F[spring] = k (0.79 m)
which counters the weight of the mass,
F[weight] = (0.46 kg) g = 4.508 N
so that by Newton's second law,
F[spring] - F[weight] = 0 ⇒ k = (4.508 N) / (0.79 m) ≈ 5.7 N/m
b. Using the same equation as before, we now have
F[weight] = (0.75 kg) g = 7.35 N
so that
(5.7 N/m) x - 7.35 N = 0 ⇒ x = (7.35 N) / (5.7 N/m) ≈ 1.3 m
