Answer:
The automobile's acceleration in that time interval is -2 m/s^2
Explanation:
The acceleration is defined as the rate of change of the velocity.
The average acceleration in a given lapse of time is calculated as:
A = (final velocity - initial velocity)/time.
In this case, we have:
initial velocity = 31 m/s
final velocity = 15 m/s
time = 8 seconds.
Then the average acceleration is:
A = (15m/s - 31m/s)/8s = -2 m/s^2
<span>The two factors that act on parachutes are gravity and air resistance, which is also called drag. Gravity acts as a force to pull parachutes down to the surface of the Earth, while air resistance generates movement in the opposite direction of the falling parachute, and essentially pushes the parachute upward. hope this helps!:)</span>
Answer:
0.108 rad/s².
Explanation:
Given that
Initial angular velocity ,ωi = 0 rad/s
Final angular velocity ωf= 0.5 rev/s
We know that
1 rev/s = 6.28 rad/s
ωf= 3.14 rad/s
t= 28.9 s
We know that (if acceleration is constant)
ωf=ωi + α t
α=Angular acceleration
3.14 = 0 + α x 28.9
Therefore the acceleration will be 0.108 rad/s².
Therefore the answer will be 0.108 rad/s².
Answer:
a) v = 2.4125 m / s , b) Em_{f} / Em₀ = 0.89
Explanation:
a) This is an inelastic crash problem, the system is made up of the four carriages, so the forces during the crash are internal and the moment is conserved
Initial
p₀ = m v₁ + 3 m v₂
Final
= (4 m) v
p₀ =p_{f}
m (v₁ + 3 v₂) = 4 m v
v = (v₁ +3 v₂) / 4
Let's calculate
v = (3.86 + 3 1.93) / 4
v = 2.4125 m / s
b) the initial mechanical energy is
Em₀ = K₁ + 3 K₂
Em₀ = ½ m v₁² + ½ 3m v₂²
The final mechanical energy
= K
Em_{f} = ½ 4 m v²
The fraction of energy lost is
Em_{f} / Em₀ = ½ 4m v² / ½ m (v₁² +3 v₂²)
Em_{f} / Em₀ = 4 v₂ / (v₁² + 3 v₂²)
Em_{f} / Em₀ = 4 2.4125² / (3.86² + 3 1.93²)
Em_{f} / em₀ = 23.28 / 26.07
Em_{f} / Em₀ = 0.89