Answer:
(A) V = 9.89m/s
(B) U = -2.50m/s
(C) ΔK.E = –377047J
(D) ΔK.E = –257750J
Explanation:
The full solution can be found in the attachment below. The east has been chosen as the direction for positivity.
This problem involves the principle of momentum conservation. This principle states that the total momentum before collision is equal to the total momentum after collision. This problem is an inelastic kind of collision for which the momentum is conserved but the kinetic energy is not. The kinetic energy after collision is always lesser than that before collision. The balance is converted into heat by friction, and also sound energy.
See attachment below for full solution.
Answer:
the mass of the body is 0.02 kg.
Explanation:
Given;
relative density of the oil,
= 0.875
mass of the object in oil,
= 0.013 kg
mass of the object in water,
= 0.012 kg
let the mass of the object in air = 
weight of the oil, 
weight of the water, 
The relative density of the oil is given as;

Therefore, the mass of the body is 0.02 kg.
Answer:
its 1/2 the mass of the object times by its velocity ^ 2
To answer the following questions for this specific problem:
a. 11.48 secs
b. Vp = a*t*3.6 =
3*11.48*3.6 = 124.0 km/h
<span>c. 9.1 secs. </span>
I am hoping that this answer has satisfied your query about
and it will be able to help you.
Answer:
F = 63N
Explanation:
M= 1.5kg , t= 2s, r = (2t + 10)m and
Θ = (1.5t² - 6t).
magnitude of the resultant force acting on 1.5kg = ?
Force acting on the mass =
∑Fr =MAr
Fr = m(∇r² - rθ²) ..........equation (i)
∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)
The horizontal path is defined as
r = (2t + 10)
dr/dt = 2, d²r/dt² = 0
Angle Θ is defined by
θ = (1.5t² - 6t)
dθ/dt = 3t, d²θ/dt² = 3
at t = 2
r = (2t + 10) = (2*(2) +10) = 14
but dr/dt = 2m/s and d²r/dt² = 0m/s
θ = (1.5(2)² - 6(2) ) = -6rads
dθ/dt =3(2) - 6 = 0rads
d²θ/dt = 3rad/s²
substituting equation i into equation ii,
Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)
∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]
∑F = 1.5(14*3+0) = 63N
F = √(Fr² +FΘ²) = √(0² + 63²) = 63N