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nalin [4]
3 years ago
15

A woman fires a rifle with barrel length of 0.5400 m. Let (0, 0) be where the 125 g bullet begins to move, and the bullet travel

s in the +x-direction. The force exerted by the expanding gas on the bullet is (16,000 + 10,000x − 26,000x2) N, where x is in meters. A) Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.
B) If the barrel is 1.05 m long, how much work is done?
Physics
1 answer:
dybincka [34]3 years ago
3 0

Answer:

Explanation:

Given that,

Length of barrel =0.54m

Mass of bullet=125g=0.125kg

Force extend

F=16,000+10,000x-26,000x²

a. Work done is given as

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=0.54m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=0.54m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=0.54m

W= 16,000(0.54) + 5000(0.54²) - 8666.667(0.54³) +0+0-0

W=8640+1458-1364.69

W=8733.31J

The workdone by the gas on the bullet is 8733.31J

b. Work done is given as

Work done when the length=1.05m

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=1.05m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=1.05m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=1.05mm

W= 16,000(1.05) + 5000(1.05²) - 8666.667(1.05³) +0+0-0

W=16800+5512.5-10032.75

W=12,279.75J

The workdone by the gas on the bullet is 12,279.75J

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sveticcg [70]

Answer:

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(b) 8.36×10⁻³ W/m²

Explanation:

The intensity of sound from an isotropic point source, with distance L is given as

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Where I = intensity of sound, P = Power from the source, L = length, π = pie.

(a)

1.4 m from the source.

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Given: P = 0.71 W, L = 1.4 m, π = 3.14.

Substitute into equation 1

I = 0.71/(4×3.14×1.4²)

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(b) 2.6 m from the source.

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Substitute into equation 1

I = 0.71/(4×3.14×2.6²)

I = 0.71/84.9056

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5 0
3 years ago
In a physics lab experiment, a compressed spring launches a 24 g metal ball at a 35o angle above the horizontal. Compressing the
Levart [38]

Answer:

k = 45.95 N/m

Explanation:

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Therefore,

v_{o}^{2} = \frac{(5.3\ m)(9.81\ m/s^{2})}{Sin\ 2(35^{o})}\\

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Kinetic\ Energy\ Gained\ By\ Ball = Elastic\ Potential\ Energy\ Stored\ in \ Spring\\\frac{1}{2}mv_{o}^{2} = \frac{1}{2}kx^{2}\\\\k = \frac{mv_{o}^{2}}{x^2} \\

where,

k = spring constant = ?

x = compression = 17 cm = 0.17 m

m = mass of ball = 24 g = 0.024 kg

Therefore,

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4 0
3 years ago
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So we can write:

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W = -36.733 J is the work performed on the car (negative because its direction is opposite to the motion of the car)

K_i = 66,120 J is the initial kinetic energy of the car

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The kinetic energy of the car can be also written as

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where:

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v is its final speed

Solving, we find

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.4 m/s

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