Push is in, pull is out
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Contact Forces
Frictional Force
Tension Force
Normal Force
Air Resistance Force
Applied Force
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Action-at-a-Distance Forces
Gravitational Force
Electrical Force
Magnetic forces
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Very bottom, last part
First drawing: Arrow down above box
Second drawing: Arrow up from below box
<h2>
Value of g on the distant planet is 16 m/s²</h2>
Explanation:
We have equation of motion v² = u² + 2as
Here initial velocity, u = 2 m/s
Final velocity, v = 10 m/s
Displacement, s = 3 m
We need to find acceleration, a.
Substituting
v² = u² + 2as
10² = 2² + 2 x a x 3
6a = 96
a = 16 m/s²
Value of g on the distant planet is 16 m/s²
Answer:
14.6 m/s
Explanation:
Momentum is conserved in the north:
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
After the collision, they stick together, so v₁ = v₂ = v.
m₁ u₁ + m₂ u₂ = (m₁ + m₂) v₂
(980 kg) (0 m/s) + (1500 kg) u = (980 kg + 1500 kg) vᵧ
1500 u = 2480 vᵧ
Momentum is conserved in the east:
m₁ u₁ + m₂ u₂ = (m₁ + m₂) v₂
(980 kg) (22.3 m/s) + (1500 kg) (0 m/s) = (980 kg + 1500 kg) vₓ
21854 = 2480 vₓ
vₓ = 8.81 m/s
The angle of v is 45.0°, so vᵧ = vₓ.
1500 u = 2480 (8.81 m/s)
u = 14.6 m/s
Answer:
0 m/s²
Explanation:
Draw a free body diagram (see attached).
Sum of the forces normal to the incline:
∑F = ma
N − mg cos θ = 0
N = mg cos θ
Sum of the forces parallel to the incline:
∑F = ma
mg sin θ − f = ma
mg sin θ − Nμ = ma
Substituting:
mg sin θ − (mg cos θ) μ = ma
g sin θ − g μ cos θ = a
a = g (sin θ − μ cos θ)
Given that g = 9.8 m/s², θ = 30°, and μ = 1/√3:
a = (9.8 m/s²) (sin 30° − 1/√3 cos 30°)
a = (9.8 m/s²) (1/2 − 1/2)
a = 0 m/s²
The acceleration down the incline is 0 m/s².
Answer:
The answer is Constructive interference
Explanation:
cause you said so