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alexira [117]
3 years ago
8

What average force is needed to accelerate a 9.20-gram pellet from rest to 125 m/s over a distance of 0.800 m along the barrel o

f a rifle?
Physics
1 answer:
aleksklad [387]3 years ago
8 0
Given:
u = 0, initial velocity
v = 125 m/s, final velocity
s = 0.0800 m, distance traveled
9.20 g, the mass of the pellet

If the acceleration is a, then
0² + 2*(a m/s²)*(0.800 m) = (125 m/s)²
1.6a = 15625
a = 9765.625 m/s²

Calculate the force.
F = (9.20 x 10⁻³ kg)*(9765.625 m/s²) = 98.84 N

Answer: 98.8 N (nearest tenth)
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A merry-go-round has a small box placed on it at a distance R' from its axis of
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The time after it starts spinning, that the box will slip off is 0.1 s.

<h3>Apply the principle of conservation of angular momentum</h3>

I₁ω₁ - I₂ω₂ = 0

I₁ω₁  =  I₂ω₂

where;

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The final angular speed when the box slides off;

ω₂ = I₁ω₁ / I₂

ω₂  = [0.5(m₁ + m₂)(R + r)²] / (0.5MR²)

ω₂  = [0.5(30 + 0.3)(3.1 + 1.4)²] / (0.5 x 30 x 3.1²)

ω₂  = 2.13 rad/s

Time taken for the for box to slide off;

τ = Iα

τ = I(ω/t)

τ = (Iω)/t

t =  (Iω)/τ

t = (0.5 x 0.3 x 1.4² x 2.13)/6

t = 0.1 s

Thus, the time after it starts spinning, that the box will slip off is 0.1 s.

Learn more about moment of inertia here: brainly.com/question/3406242

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Determine the lamp wattage required to obtain an illumination level of 50 f-c over a 100 ft2 area if a fixture is used with a CU
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Lamp Wattage is utilized with a CU of 0.75 and 80 percent of the available light reaches the work surface, the remaining 20 percent is absorbed by walls and other objects in the space, resulting in an illumination level of 50 f-c over a 100 ft2 area

The term "lumen" refers to the emission of "luminous flux," which is a measurement of the total amount of visible light emitted by a source in a certain amount of time. The illumination level over a 100 ft2 area will be 50 f-c Lamp Wattage with a CU of 0.75 used, with 80 percent of the available light reaching the work surface and the remaining 20 percent being absorbed by walls and other objects in the room.
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