Answer:
1. external force;not change
2. all of the above
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Answer:
(a). The initial velocity is 28.58m/s
(b). The speed when touching the ground is 33.3m/s.
Explanation:
The equations governing the position of the projectile are
where is the initial velocity.
(a).
When the projectile hits the 50m mark, ; therefore,
solving for we get:
Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that
which gives
(b).
The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,
the vertical component of the velocity is
which gives a speed of
#1
As per kinematics we know that
here we know that
now we will have
Now by Newton's II law
#2
part a)
Impulse = change in momentum
given that
m = 0.153 kg
v = 25.7 m/s
now we have
Part b)
As per Newton's II law we know that
as we know that
Answer:
L = 4.58 x 10⁴ kg.m²/s
Explanation:
The angular momentum is given by the formula:
L = mvr
but, v = rω
Therefore,
L = mr²ω
where,
L = Angular Momentum of the person = ?
m = mass of person = 99 kg
r = radius of earth = 6.37 x 10⁶ m
ω = Angular Speed of Earth's Rotation = θ/t
Since, earth completes 1 rotation in 1 day. Hence,
ω = (2π rad/1 day)(1 day/24 h)(1 h/3600 s)
ω = 7.27 x 10⁻⁵ rad/s
Therefore,
L = (99 kg)(6.37 x 10⁶ m)²(7.27 x 10⁻⁵ rad/s)
<u>L = 4.58 x 10⁴ kg.m²/s</u>