Answer: D.) 39,200 J
Via the equation of potential energy PE = mgh where m is mass, g is the average gravity on earth and h is the height. In this case m = 400 kg, g = 9.8, h = 10 m thus:
P.E.= 39,200 Joules
Pleaseeee help I really need to pass to graduate
1 answer:
You might be interested in
Answer:
0.25m/s
Explanation:
Given parameters
m₁ = 5kg
v₁ = 1.0m/s
m₂ = 15kg
v₂ = 0m/s
Unknown:
velocity after collision = ?
Solution:
Momentum before collision and after collision will be the same. For inelastic collision;
m₁v₁ + m₂v₂ = v(m₁ + m₂)
Insert parameters and solve for v;
5 x 1 + 15 x 0 = v (5 + 15 )
5 = 20v
v = = 0.25m/s
Answer:
The tension in each half of the rope, is approximately 4,908.8 N
Explanation:
The mass of the acrobat, m = 60 kg
The length of the rope, l = 10 m
The extent by which the center dropped = 30 cm = 0.3 m
Let, 'T' represent the tension in each half of the rope
Weight, W = Mass, m × The acceleration due to gravity, g
∴ W = m × g
The acceleration due to gravity, g ≈ 9.8 m/s²
∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N
The angle the dropped rope makes with the horizontal, θ is given as follows;
θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°
At equilibrium, the sum of vertical forces, = 0
The vertical component of the tension, , in each half of the rope is given as follows;
= T × sin(θ)
∴ = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)
Plugging in the values, with θ = arctan(0.06) for accuracy, we get;
588 N + 2 × T × sin(arctan(0.06) = 0
∴ 2 × -T × sin(arctan(0.06) = 588 N
-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N
The tension in each half of the rope, T ≈ 4,908.8 N.
