Answer:
= 3,126 m / s
Explanation:
In a crash exercise the moment is conserved, for this a system formed by all the bodies before and after the crash is defined, so that the forces involved have been internalized.
the car has a mass of m = 1.50 kg a speed of v1 = 4.758 m / s and the mass of the train is M = 3.60 kg and its speed v2 = 2.45 m / s
Before the crash
p₀ = m v₁₀ + M v₂₀
After the inelastic shock
= m
+ M
p₀ = 
m v₀ + M v₂₀ = m
+ M
We cleared the end of the train
M
= m (v₁₀ - v1f) + M v₂₀
Let's calculate
3.60 v2f = 1.50 (2.15-4.75) + 3.60 2.45
= (-3.9 + 8.82) /3.60
= 1.36 m / s
As we can see, this speed is lower than the speed of the car, so the two bodies are joined
set speed must be
m v₁₀ + M v₂₀ = (m + M)
= (m v₁₀ + M v₂₀) / (m + M)
= (1.50 4.75 + 3.60 2.45) /(1.50 + 3.60)
= 3,126 m / s
Answer:
10,000kgm/s
Explanation:
Since we not told what to look for, we can as well find the momentum of the car.
momentum = mas * velocity
Given
Mass of the car = 2000kg
velocity = 5m/s
Substitute into the formula
Momentum = 2000 * 5
Momentum = 10000kgm/s
Hence the momentum of the car is 10,000kgm/s
AnswerThe current through a toaster connected to a 120-V source is 8.0 A. What power is given off by the toaster? p= 8.0 A : 120V (p=960 W Page 2 7. The resistance of an electric stove eleinent at operating temperature is 11 12.
Explanation:
hope this helps =)
The force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.
<h3>Force exerted outside the wheel</h3>
The force exerted on the outside of the wheel can be determined by applying the principle of conservation of angular momentum as shown below.
∑τ = 0
- Let the distance traveled by the load = 1.5 m
- Let the radius of the wheel or position of the force = 0.45 m
∑τ = R(mg) - r(F)
rF = R(mg)
0.45F = 1.5(21,200 x 9.8)
F = 6.925 x 10⁵ N.
Thus, the force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.
Learn more about angular momentum here: brainly.com/question/7538238