<u>Answer:</u>
The force F applied to the handle = 330.03 N
<u>Explanation:</u>
The force can be resolved in to two, horizontal component and vertical component. If θ is the angle between horizontal and applied force we have
Horizontal component of force = F cos θ
Vertical component of force = F sin θ
In this problem normal force exerted on the suitcase is 160 N, that is vertical component of force = 160 N and angle θ = 29⁰.
So, F sin 29 = 160
F = 330.03 N
The force F applied to the handle = 330.03 N
The anwser is 1.475124 horsepower.
T1=S/V1T2=Ttot-T1=(2S/Vtot)-S/V1V2=S/T2 = S / (2S/Vtot-S/V1)Simplify in V2 = V1Vtot/(2V1-Vtot) = 48 * 90 / 6 = 720 km/hr
Answer: 1
an object positioned at some height in a gravitational field
Explanation:
Gravitational potential energy of an object is the energy stored due to position of the object or position at certain height relative to zero position.
Gravitational potential energy can also be expressed as object position at some height above or below zero position in a gravitational field
I think 1 and 2 make sense. But 1 make more sense than 2
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