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Vaselesa [24]
3 years ago
7

A pivot at the 35cm mark supports a meter stick. It is brought into a balance by a 0.85N weight suspended at the 5.5cm mark.comp

ute the weight of the meter stick.
Physics
1 answer:
Arte-miy333 [17]3 years ago
8 0

Answer:

2.52N

Explanation:

according to law of moments ,

F1L1=F2L2 ,where the forces are in equilibrium .

0.85(44.5)= 15F2

that is , taking their distances from 50cm point

37 .825=15F2

37.825÷15= F2

2.52N = F2

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A 540 kg satellite moves through deep space with a speed of 27 m/s. A booster rocket on the satellite fires for 1.4 s, giving a
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Question 23
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If the gymnast mass were doubled, her height (h) from the top of the board would be as follows,

с  Stay the same

Explanation:

  • The Mass of an object or body does not affect the acceleration due to gravity in any kind of way.
  • Light weight objects accelerate more slowly than the heavy objects because when the forces other than the gravity also plays a major role.
  • Mass increases of a body when an object has higher velocity or the speed.
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3 0
3 years ago
Unpolarized light of intensity I0 = 950 W/m2 is incident upon two polarizers. The first has its polarizing axis vertical, and th
Ket [755]

Answer:

Intensity of the light (first polarizer) (I₁) = 425 W/m²

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

Explanation:

Given:

Unpolarized light of intensity (I₀) = 950 W/m²

θ = 65°

Find:

a. Intensity of the light (first polarizer)

b. Intensity of the light (second polarizer)

Computation:

a. Intensity of the light (first polarizer)

Intensity of the light (first polarizer) (I₁) = I₀ / 2

Intensity of the light (first polarizer) (I₁) = 950 / 2

Intensity of the light (first polarizer) (I₁) = 425 W/m²

b. Intensity of the light (second polarizer)

Intensity of the light (second polarizer) (I₂) = (I₁)cos²θ

Intensity of the light (second polarizer) (I₂) = (425)(0.1786)

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

5 0
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