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➷ Earth's gravity is approximately 9.81
weight = mass x gravity
weight = 4.6 x 9.81
weight = 45.126
Answer is B. 45N
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the answer should be:
When the buoyant force is equal to the force of gravity
Answer:
(a) The range of the projectile is 31,813.18 m
(b) The maximum height of the projectile is 4,591.84 m
(c) The speed with which the projectile hits the ground is 670.82 m/s.
Explanation:
Given;
initial speed of the projectile, u = 600 m/s
angle of projection, θ = 30⁰
acceleration due to gravity, g = 9.8 m/s²
(a) The range of the projectile in meters;

(b) The maximum height of the projectile in meters;

(c) The speed with which the projectile hits the ground is;

Answer:
t = 5.05 s
Explanation:
This is a kinetic problem.
a) to solve it we must fix a reference system, let's use a fixed system on the floor where the height is 0 m
b) in this system the equations of motion are
y = v₀ t + ½ g t²
where v₀ is the initial velocity that is v₀ = 0 and g is the acceleration of gravity that always points towards the center of the Earth
e) y = 0 + ½ g t²
t = √ (2y / g)
t = √(2 125 / 9.8)
t = 5.05 s
Complete question is;
A rocket ship starts from rest and turns on its forward booster rockets, causing it to have a constant acceleration of 4 m/s² rightward. After 3s, what will be the velocity of the rocket ship?
Answer:
v = 12 m/s
Explanation:
We are given;
Initial velocity; u = 0 m/s (because ship starts from rest)
Acceleration; a = 4 m/s²
Time; t = 3 s
To find velocity after 3 s, we will use Newton's first equation of motion;
v = u + at
v = 0 + (4 × 3)
v = 12 m/s