Answer: C
Explanation:
In collision, whether elastic or inelastic collisions, momentum is always conserved. That is, the momentum before collision will be equal to the momentum after collision.
Change in momentum of the system will be momentum after collision minus total momentum before collision.
Since momentum is a vector quantity, the direction will also be considered.
Momentum = MV - mU
Let
M = 800 kg is going north
at V = 20 m/s and the other car
m= 800 kg is going south
at U = 10m/s.
Substitute all the parameters into the formula
Momentum = (800 × 20) - (800 × 10)
= 8000 kgm/s
The final momentum after collision will also be equal to 8000 kgm/s
Change in momentum = 8000 - 8000
Change in momentum = 0
All you need to know is that velocity is the derivative with respect to time of position.
Therefore: ds/dt = -32t +10
To know the velocity at t=2s you just need to plug t=2 for t in the equation above.
Answer:
The intensity of sound at rock concert is 10¹⁰ greater than that of a whisper.
Explanation:
The intensity of sound is given by;
where;
I is the intensity of the sound
I₀ is the threshold of sound intensity = 1 x 10⁻¹² W/m²
The intensity of sound at a rock concert
The intensity of sound of a whisper
Thus, the intensity of sound at rock concert is 10¹⁰ greater than that of a whisper.
The new rotational energy of the system(RE) = 
How this is calculated?
- Distance of each astronaut from centre of mass(m)=d/2
- Moment of inertia of system
- So, angular speed,ω =
=
- Rotational energy of system=
where, ω= angular speed
What is rotational energy?
- Rotational energy or angular kinetic energy is kinetic energy due to the rotation of an object and is part of its total kinetic energy.
- Looking at rotational energy separately around an object's axis of rotation, the following dependence on the object's moment of inertia is observed
To know more about rotational energy , refer:
brainly.com/question/13623190
#SPJ4
Answer:
W₃ = 3310.49 J
, W3 = 3310.49 J
Explanation:
We can solve this exercise in parts, the first with acceleration, the second with constant speed and the third with deceleration. Therefore it is work we calculate it in these three sections
We start with the part with acceleration, the distance traveled is y = 5.90 m and the final speed is v = 2.30 m / s. Let's calculate the acceleration with kinematics
v2 = v₀² + 2 a₁ y
as they rest part of the rest the ricial speed is zero
v² = 2 a₁ y
a₁ = v² / 2y
a₁ = 2.3² / (2 5.90)
a₁ = 0.448 m / s²
with this acceleration we can calculate the applied force, using Newton's second law
F -W = m a₁
F = m a₁ + m g
F = m (a₁ + g)
F = 69 (0.448 + 9.8)
F = 707.1 N
Work is defined by
W₁ = F.y = F and cos tea
As the force lifts the man, this and the displacement are parallel, therefore the angle is zero
W₁ = 707.1 5.9
W₁ = 4171.89 J W3 = 3310.49 J
Let's calculate for the second part
the speed is constant, therefore they relate it to zero
F - W = 0
F = W
F = m g
F = 60 9.8
F = 588 A
the job is
W² = 588 5.9
W2 = 3469.2 J
finally the third part
in this case the initial speed is 2.3 m / s and the final speed is zero
v² = v₀² + 2 a₂ y
0 = vo2₀² + 2 a₂ y
a₂ = -v₀² / 2 y
a₂ = - 2.3²/2 5.9
a2 = - 0.448 m / s²
we calculate the force
F - W = m a₂
F = m (g + a₂)
F = 60 (9.8 - 0.448)
F = 561.1 N
we calculate the work
W3 = F and
W3 = 561.1 5.9
W3 = 3310.49 J
total work
W_total = W1 + W2 + W3
W_total = 4171.89 +3469.2 + 3310.49
w_total = 10951.58 J
