Answer:
4960 N
Explanation:
First, find the acceleration.
Given:
v₀ = 6.33 m/s
v = 2.38 m/s
Δx = 4.20 m
Find: a
v² = v₀² + 2aΔx
(2.38 m/s)² = (6.33 m/s)² + 2a (4.20 m)
a = -4.10 m/s²
Next, find the force.
F = ma
F = (1210 kg) (-4.10 m/s²)
F = -4960 N
The magnitude of the force is 4960 N.
Answer:
1.28 x 10^4 N
Explanation:
m = 1500 kg, h = 450 km, radius of earth, R = 6400 km
Let the acceleration due to gravity at this height is g'
g' / g = {R / (R + h)}^2
g' / g = {6400 / (6850)}^2
g' = 8.55 m/s^2
The force between the spacecraft and teh earth is teh weight of teh spacecraft
W = m x g' = 1500 x 8.55 = 1.28 x 10^4 N
Answer:
c. selenium
Explanation:
sulfur and selenium are in the same group
Answer:
a = 1.764m/s^2
Explanation:
By Newton's second law, the net force is F = ma.
The equation for friction is F(k) = F(n) * μ.
In this case, the normal force is simply F(n) = mg due to no other external forces being specified
F(n) = mg = 15kg * 9.8 m/s^2 = 147N.
F(k) = F(n) * μ = 147N * 0.18 = 26.46N.
Assuming the object is on a horizontal surface, the force due to gravity and the normal force will cancel each other out, leaving our net force as only the frictional one.
Thus, F(net) = F(k) = ma
26.46N = 15kg * a
a = 1.764m/s^2
