Answer:
i) 3.514 s, ii) 5.692 m/s
Explanation:
i) We can use Newton's second law of motion to find out how long does it take for the Eagle to touch down.
as the equation says for free-falling
h = ut +0.5gt^2
Here, h = 10 m, g = acceleration due to gravity = 1.62 m/s^2( on moon surface)
initial velocity u = 0
10 = 0.5×1.62t^2
t = 3.514 seconds
Therefore, it takes t = 3.514 seconds for the Eagle to touch down.
ii) use Newton's 1st equation of motion to calculate the velocity of the lunar module when it hits the surface of the moon
v = u + gt
v = 0+ 1.62×3.514
v= 5.692 m/s
Answer: a = 4 m/s²
Explanation:
a = Δv/t = (30 - 18) / 3 = 4 m/s²
Answer:
Force of friction is (-30 N).
Explanation:
The force applied on the box across the floor is 30 N.
The force of gravity is (-8 N) and the the normal force is 8 N.
It is based on Newton's third law of motion. Newton's third law of motion states that the force acting on object 1 to object 2 is equal in magnitude of the force from object 2 to 1 but in opposite direction.
Here there is force of 30 N is applied in horizontal direction. The frictional force act in opposite direction. So, the force of friction is -30 N so that box across the floor.
Answer:
392 N
Explanation:
Draw a free body diagram of the rod. There are four forces acting on the rod:
At the wall, you have horizontal and vertical reaction forces, Rx and Ry.
At the other end of the rod (point X), you have the weight of the sign pointing down, mg.
Also at point X, you have the tension in the wire, T, pulling at an angle θ from the -x axis.
Sum of the moments at the wall:
∑τ = Iα
(T sin θ) L − (mg) L = 0
T sin θ − mg = 0
T = mg / sin θ
Given m = 20 kg and θ = 30.0°:
T = (20 kg) (9.8 m/s²) / (sin 30.0°)
T = 392 N
Answer:
1400 m
Explanation:
Given:
v₀ = 175 m/s
v = 105 m/s
t = 10.0 s
Find: Δx
Δx = ½ (v + v₀) t
Δx = ½ (105 m/s + 175 m/s) (10.0 s)
Δx = 1400 m
