Answer:
force exerted by Jet pack is 63 N on the astronaut
Explanation:
As per Newton's II law we know that
Rate of change in momentum of the system is net force on it
so we have
so we have
so we have
Since he used jet pack to return back towards the spaceship
so here we will have
force exerted by Jet pack is 63 N on the astronaut
The tension in rope C is determined as 122.23 N.
<h3>Tension in rope C</h3>
The tension in rope C is calculated as follows;
B² = A² + C²
C² = B² - A²
where;
- C is tension in rope C
- A is tension in rope A
- B is tension in rope B
C² = (752²) - (742²)
C² = 14,940
C = √14,940
C = 122.23 N
Thus, the tension in rope C is determined as 122.23 N.
Learn more about tension here: brainly.com/question/24994188
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I have no idea... Sorry :D
The effective mass of the system is 12 kg.
<u>Explanation:</u>
- The effective mass of the system is given by the expression,
= mₓ + 1/2 m₁ + 2/5 m₂
- In the above equations the subscript x denotes block B and the subscript 2 denotes the mass of the solid sphere and the subscript 1 denotes the pulley.
- Expressing the above functions with the data given below. we have,
= 6 kg + 1/2 (4 kg) + 2/5 (10 kg)
= 12 kg.
with the findings above we can conclude that the effective mass of the system is 12 kg
Answer:
This is an inelastic collision. This means, unfortunately, that KE cannot save you, at least in the problem's current form.
Let's see what conservation of momentum in both directions does ya:
Conservation in the x direction:
Only 1 object here has a momentum in the x direction initally.
m1v1i + 0 = (m1 + m2)(vx)
3.09(5.10) = (3.09 + 2.52)Vx
Vx = 2.81 m/s
Explanation:
Conservation in the y direction:
Again, only 1 object here has initial velocity in the y:
0 + m2v2i = (m1 +m2)Vy
(2.52)(-3.36) = (2.52 + 3.09)Vy
Vy = -1.51 m/s
++++++++++++++++++++
Now that you have Vx and Vy of the composite object, you can find the final velocity by doing Vf = √Vx^2 + Vy^2)
Vf = √(2.81)^2 + (-1.51)^2
Vf = 3.19 m/s
