Answer:
The instantaneous velocity of the rocket the moment before it hits the ground is 50 m/s.
Explanation:
Given;
initial velocity of the rocket, u = 50 m/s
Determine the maximum height reached by the rocket.
at maximum height reached by the rocket, the final velocity, v = 0
v² = u² -2gh
0 = 50² - 2(9.8)h
19.6h = 2500
h = 2500 / 19.6
h = 127.55 m
At maximum height, the time to reach ground is given by;
h = ¹/₂gt²
Before the rocket hits the ground the final velocity will be maximum;
v = u + gt
v = 0 + 9.8 x 5.1
v = 50 m/s
Therefore, the instantaneous velocity of the rocket the moment before it hits the ground is 50 m/s.
48J
Explanation:
Given parameters:
Weight of the book = 16N
Height of the shelf = 3m
Unknown:
Work done to raise the book = ?
Solution:
Work done is defined as the product of force and distance. The force here is the weight of the book which signifies that it was carried in the vicinity of gravitational force.
Work done = Force x distance = weight x distance
Weight is a type of force;
Input the parameters;
Work done = 16 x 3 = 48J
The unit of workdone is in Joules
learn more:
Work done brainly.com/question/9100769
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Answer:
a) 
, b) 
Explanation:
The problem is asking the rocket velocity and acceleration at t = 6 s.
a) The general equation of the rocket is:
b) The acceleration experimented by the rocket is:
Answer:
The maximum height that a cannonball fired at 420 m/s at a 53.0° angles is 5740.48m.
hmax = 5740.48 m
Explanation:
This is an example of parabolic launch. A cannonball is fired on flat ground at 420 m/s at a 53.0° angle and we have to calculate the maximum height that it reach.
V₀ = 420m/s and θ₀ = 53.0°
So, when the cannonball is fired it has horizontal and vertical components:
V₀ₓ = V₀ cos θ₀ = (420m/s)(cos 53°) = 252.76 m/s
V₀y = V₀ cos θ₀ = (420m/s)(cos 53°) = 335.43m/s
When the cannoball reach the maximum height the vertical velocity component is zero, that happens in a tₐ time:
Vy = V₀y - g tₐ = 0
tₐ = V₀y/g
tₐ = (335.43m/s)/(9.8m/s²) = 34.23s
Then, the maximum height is reached in the instant tₐ = 34.23s:
h = V₀y tₐ - 1/2g tₐ²
hmax = (335.43m/s)(34.23s)-1/2(9.8m/s²)(34.23s)²
hmax = 11481.77m - 5741.29m
hmax = 5740.48m
Answer:
40
Explanation:
Mechanical advantage = effort arm / load arm
MA = 20 cm / 0.5 cm
MA = 40
