Answer:
maybeeeeeeeee launcherrr im not sureee hehe
Explanation:
A worker picks up the bag of gravel. We need to find the speed of the bucket after it has descended 2.30 m from rest. It is case of conservation of energy. So,
h = 2.3 m
So, the speed of the bucket after it has descended 2.30 m from rest is 6.71 m/s.
Answer: to only change one factor in an experiment or test
In the presence of air resistance, a watermelon is launched into the air with 100 j of kinetic energy.
Its kinetic energy is less than 100 J when it reaches its starting point. Its kinetic energy decreases as it encounters air resistance and returns to its starting point. In actuality, some of the energy has been lost because of air resistance. Since we use the ball's original height as a point of reference, there is no potential energy when the ball is in its initial state of motion, and K is its kinetic energy. This total energy is conserved if there is no air resistance, therefore when the ball returns to its starting position, its kinetic energy will remain at 100.
Learn more kinetic energy about here:
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Answer:
The inductance of solenoid A is twice that of solenoid B
Explanation:
The inductance of a solenoid L is given by
L = μ₀n²Al where n = turns density, A = cross-sectional area of solenoid and l = length of solenoid.
Given that d₁ = 2d₂ and l₂ = 2l₁ and d₁ and d₂ are diameters of solenoids A and B respectively. Also, l₁ and l₂ are lengths of solenoids A and B respectively.
Since we have a cylindrical solenoid, the cross-section is a circle. So, A = πd²/4.
Let L₁ and L₂ be the inductances of solenoids A and B respectively.
So L₁ = μ₀n²A₁l₁ = μ₀n²πd₁²l₁/4
L₂ = μ₀n²A₂l₂ = μ₀n²πd₂²l₂/4
Since d₁ = 2d₂ and l₂ = 2l₁, sub
L₁/L₂ = μ₀n²πd₁²l₁/4 ÷ μ₀n²πd₂²l₂/4 = d₁²/d₂² × l₁/l₂ = (2d₂)²/d₂² × l₁/2l₁ = 4d₂²/d₂² × l₁/2l₁ = 4 × 1/2 = 2
L₁/L₂ = 2
L₁ = 2L₂
So, the inductance of solenoid A is twice that of solenoid B
