A ball is released from a hot air balloon moving downward with a velocity of -10.0 meters/second and a height of 1,000 meters. H
2 answers:
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Answer:
v₁₀ = 1.90 m / s
Explanation:
In this exercise we are given the maximum height data, with energy we can know how fast the body came out
Final mechanical energy, maximum height
= U = m g h
Initial mechanical energy, in the lower part of the track
Em₀ = K = ½ m v²
Em=
½ m v² = m g h
v = √ 2gh
Now we can use the moment to find the speed with which objects collide
The large object has a mass M = 5.41 kg a velocity starts v₁₀, the small object has a mass m = 1.68 kg an initial velocity of zero v₂₀ = 0 and final velocity v
Initial before the crash
p₀ = M v₁₀ + 0
Final after the crash
= M v1f + m v
p₀ =
M v₁₀ = M + m v
As the shock is elastic the kinetic energy is conserved
K₀ =
½ M v₁₀² = ½ M ² + ½ m v²
Let's write the system of equations
M v₁₀ = M + m v
M v1₁₀² = M ² + m v²
We cleared v1f in the first we replaced in the second
= (M v₁₀ - mv) / M
M v₁₀² = M (M v₁₀ - mv)² / M² + m v²
M v₁₀² = 1 / M (M² v₁₀² - 2mM v v₁₀ + m² v²) +m v²
v₁₀² (M - M) + 2 m v v₁₀ - v² (m2 + m) / M = 0
2 m v₁₀ - v (m + 1) m/ M = 0
v₁₀ = v (m +1) / (2M)
Let's substitute the value of v
v1₁₀= √ (2gh) (m +1) / (2M)
Let's calculate
v₁₀ = √ (2 9.8 3) (1+ 1.68) / (2 5.41)
V₁₀ = 7.668 (2.68) / 10.82
v₁₀ = 1.90 m / s
Answer:
The ratio of their orbital speeds are 5:4.
Explanation:
Given that,
Mass of A = 5 m
Mass of B = 7 m
Radius of A = 4 r
Radius of B = 7 r
The orbital speed of satellite A,
......(I)
The orbital speed of satellite B,
......(I)
We need to calculate the ratio of their orbital speeds
Using equation (I) and (II)
Put the value into the formula
Hence, The ratio of their orbital speeds are 5:4.