A 0.55 kg projectile is launched from the edge of a cliff with an initial kinetic energy of 1550 J and at its highest point is 1
1 answer:
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The value of the force, F₀, at equilibrium is equal to the horizontal
component of the tension in string 2.
Response:
- The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>
Given:
The weight of the rod = The sum of the vertical forces in the strings
Therefore;
M·g = T₂·cos(37°) + T₁
The weight of the rod is at the middle.
Taking moment about point (2) gives;
M·g × L = T₁ × 2·L
Therefore;
Which gives;
F₀ = T₂·sin(37°)
Which gives;
- F₀ ≈ <u>0.377·M·g</u>
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Learn more about equilibrium of forces here:
Answer:
(a) 490 N on earth
(b) 80 N on earth
(c) 45.4545 kg on earth
(d) 270.27 kg on moon
Explanation:
We have given 1 kg = 9.8 N = 2.2 lbs on earth
And 1 kg = 1.6 N = 0.37 lbs on moon
(a) We have given mass of the person m = 50 kg
As it is given that 1 kg = 9.8 N
So 50 kg = 50×9.8 =490 N
(b) Mass of the person on moon = 50 kg
As it is given that on moon 1 kg = 1.6 N
So 50 kg = 50×1.6 = 80 N
(c) We have given that weight of the person on the earth = 100 lbs
As it is given that 1 kg = 2.2 lbs on earth
So 100 lbs = 45.4545 kg
(d) We have given weight of the person on moon = 100 lbs
As it is given that 1 kg = 0.37 lbs
So 100 lbs