Explanation:
Since the balloon is not accelerating means that the net force on the balloon is zero. This implies that the weight of balloon must be equal to the buoyant force on balloon.
Hence, the buoyant force equals the weight of air displaced by the balloon, also 20,000 N.
Weight of the air displaced = density of air × volume
The density of air at 1 atm pressure and 20º C is 1.2 kg/m³
the volume V = 20,000/(1.2×9.8) = 1700 m³
According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.
To solve this problem we must apply the concept related to the conservation of energy theorem.
PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so



PART B) Replacing the values given as,




Therefore the speed of the masses would be 1.8486m/s
The initial velocity of the ball is 55.125 m/s.
<h3>Initial velocity of the ball</h3>
The initial velocity of the ball is calculated as follows;
During upward motion
h = vi - ¹/₂gt²
h = vi - 0.5(9.8)(3²)
h = vi - 44.1 ----------------- (1)
During downward motion
h = vi + ¹/₂gt²
h = 0 + 0.5(9.8)(1.5)²
h = 11.025 ----------- (2)
solve (1) and (2) together, to determine the initial velocity of the ball
11.025 = vi - 44.1
vi = 11.025 + 44.1
vi = 55.125 m/s
Thus, the initial velocity of the ball is 55.125 m/s.
Learn more about initial velocity here: brainly.com/question/19365526
#SPJ1
Answer:
1. True WA > WB > WC
Explanation:
In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.
A) Work is the product of force by distance and the cosine of the angle between them
WA = W h cos 0
WA = mg h
B) On a ramp without rubbing
Sin30 = h / L
L = h / sin 30
WB = F d cos θ
WB = F L cos 30
WB = mf (h / sin30) cos 30
WB = mg h ctan 30
C) Ramp with rubbing
W sin 30 - fr = ma
N- Wcos30 = 0
W sin 30 - μ W cos 30 = ma
F = W (sin30 - μ cos30)
WC = mg (sin30 - μ cos30) h / sin30
Wc = mg (1 - μ ctan30) h
When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum
Let's review the claims
1. True The work of gravity is the greatest and the work where there is friction is the least
2 False. The job where there is friction is the least
3 False work with rubbing is the least
4 False work with rubbing is the least
Answer:
164.87 J
Explanation:
From the question,
Work done (W) = mghcosθ........................ Equation 1
Where m = mass of the box, h = height, g = acceleration due to gravity, θ = angle to the vertical
Given: m = 25 kg, h = 2.6 meters, θ = 75°.
Constant: g = 9.8 m/s²
Substitute these value into equation 1
W = 25×9.8×2.6×cos75°
W = 164.87 J.