Answer:
a) a = - 0.106 m/s^2 (←)
b) T = 12215.1064 N
Explanation:
If
F₁ = 9*1350 N = 12150 N (→)
F₂ = 9*1365 N = 12285 N (←)
∑Fx = M*a = (M₁ +M₂)*a (→)
F₁ - F₂ = (M₁ +M₂)*a
→ a = (F₁ - F₂) / (M₁ +M₂ ) = (12150-12285)N/(9*68+9*73)Kg
→ a = - 0.106 m/s^2 (←)
(b) What is the tension in the section of rope between the teams?
If we apply ∑Fx = M*a for the team 1
F₁ - T = - M₁*a ⇒ T = F₁ + M₁*a
⇒ T = 12150 N + (9 * 68 Kg)*(0.106 m/s^2)
⇒ T = 12215.1064 N
If we choose the team 2 we get
- F₂ + T = - M₂*a ⇒ T = F₂ - M₂*a
⇒ T = 12285 N - (9 * 73 Kg)*(0.106 m/s^2)
⇒ T = 12215.1064 N
Kinetic energy has nothing to do with anything other than motion of the particle.
When a particle with velocity v collides another particle(suppose it is at rest for simplication), assuming that there is perfectly elastic collision between them, the velocity of particle which was at rest becomes mv/M ( assuming mass of particle in motion to be m and at rest to be M) from convervation of linear momentum. And all this transfer of energy happens in a fraction of seconds which is not visible to naked eyes.
Hence 1st option is correct!
Answer:
The coefficient of friction and acceleration are 0.37 and 2.2 m/s²
Explanation:
Suppose we find the coefficient of friction and the acceleration of the 100 kg block during the time that the 60 kg block remains in contact.
Given that,
Mass of block = 60 kg
Acceleration = 2.0 m/s²
Mass = 100 kg
Horizontal force = 340 N
Let the frictional force be f.
We need to calculate the frictional force
Using balance equation

Put the value into the formula



We need to calculate the coefficient of friction
Using formula of friction force




We need to calculate the acceleration of the 100 kg block
Using formula of newton's law




Hence, The coefficient of friction and acceleration are 0.37 and 2.2 m/s²
It will be moving at high speeds
Answer:
B.
Explanation:
The sound waves are compromised. It decrompesses each time which makes the sound gets lower and lower.