Answer:
The height of the cliff is 121.276 m
Explanation:
Given;
initial velocity of the projectile, v₁ = 75 m/s
final velocity of the projectile, v₂ = 90 m/s
spring compression = 5 m
Apply the law of conservation of energy;
mgh₀ + ¹/₂mv₁² = mgh₂ + ¹/₂mv₂²
gh₀ + ¹/₂v₁² = gh₂ + ¹/₂v²
gh₁ - gh₂ = ¹/₂v₂² - ¹/₂v₁²
g(h₀ - h₂) = ¹/₂ (v₂² - v₁²)
h₀ - h₂ = ¹/₂g (v₂² - v₁²)
h₀ = h(cliff) + 5m
when the projectile hits the ground, Final height, h₂ = 0
Therefore, the height of the cliff is 121.276 m
Answer:
(a) 0.833 j
(b) 2.497 j
(c) 4.1625 j
(d) 4.995 watt
Explanation:
We have given force F = 5 N
Mass of the body m = 15 kg
So acceleration
As the body starts from rest so initial velocity u = 0 m/sec
(a) From second equation of motion
For t = 1 sec
We know that work done W =force × distance = 5×0.1666 =0.833 j
(b) For t = 2 sec
We know that work done W =force × distance = 5×0.666 =3.33 j
So work done in second second = 3.33-0.833 = 2.497 j
(c) For t = 3 sec
We know that work done W =force × distance = 5×1.4985 =7.4925 j
So work done in third second = 7.4925 - 2.497 -0.833 = 4.1625 j
(d) Velocity at the end of third second v = u+at
So v = 0+0.333×3 = 0.999 m /sec
We know that power P = force × velocity
So power = 5× 0.999 = 4.995 watt
The half-meter rule (easy math) is 0.5 meters or 50 centimeters since a meter is 1 meters long, which is equivalent to 100 centimeters. Therefore, we shall apply the 50 cm rule.
A 50 cm rule's center of mass is now 25 cm away.
Additionally, according to the data, the object is pivoted at 15 cm, while the 40 g object is hung at 2 cm from the rule's beginning. Using a straightforward formula, we can compare the two situations: the distance from the pivot to the center of the mass times the mass of the 40 g object divided by 2 cm must equal the distance from the pivot to the center of the mass times mass of the 10 x g object
The result of the straightforward computation must be 52g.
Most simplified version:
the center of mass of the rule is at the 25 cm mark
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