Answer:
Explanation:
We shall apply law of conservation of momentum .
Momentum before collision = momentum after collision .
Momentum before collision = 400 kg m/s
Momentum after collision = 5 x v + 11 x 15
where v is velocity of A after the collision .
5 x v + 11 x 15 = 400
5 v = 400 - 165
5v = 235
v = 47 m /s .
Answer:
D.
boulder-sized rocks that come from the asteroid belt
Answer:
8.00 kJ
Explanation:
The first thing is to determine what quantities are there.
the mass of water = 1 000 kg
initial velocity, u = 6 m/s
final velocity, v = 4 m/s
the generator is operating at 100 % efficiency, so there is no energy loss.
The kinetic energy, Ek is converted to electrical energy, therefore Ek = electrical energy.
The kinetic energy is calculated as follows:
Ek = 1/2 mv²
= 1/2×(1 000)× (4)²
= 8 000 J/s
= 8.00 kJ Ans
Answer:
position as a function of time is y = 0.05 × cos(9.9)t
Explanation:
given data
mass = 5 kg
length = 10 cm = 0.1 m
displaced = 5 cm
to find out
position as a function of time
solution
we will apply here equilibrium that is
mass × g = k × length
put here value and find k
k = 
k = 490 N/m
and ω is
ω = 
ω = 
ω = 9.9
so here position w.r.t time is
y = 0.05 × cosωt
y = 0.05 × cos(9.9)t
so position as a function of time is y = 0.05 × cos(9.9)t
Answer:
a) a = 34.375 m / s², b) v_f = 550 m / s
Explanation:
This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.
a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed
v_f =
we substitute the values
v_f =
The initial part of the movement is carried out with acceleration
v_f = v₀ + a t
x₁ = x₀ + v₀ t + ½ a t²
the rocket starts from rest v₀ = 0 with an initial height x₀ = 0
x₁ = ½ a t²
v_f = a t
we substitute the values
x₁ = 1/2 a 16²
x₁ = 128 a
v_f = 16 a
let's write our system of equations
v_f =
x₁ = 128 a
v_f = 16 a
we substitute in the first equation
16 a =
16 4 a = 6600 - 128 a
a (64 + 128) = 6600
a = 6600/192
a = 34.375 m / s²
b) let's find the time to reach this height
x = ½ to t²
t² = 2y / a
t² = 2 5100 / 34.375
t² = 296.72
t = 17.2 s
We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part
v_f = 16 a
v_f = 16 34.375
v_f = 550 m / s