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solniwko [45]
3 years ago
7

A 7.9 g bullet leaves the muzzle of a rifle with ayvaci (na25565) – Chapter 4 Basics – balantic – (APPhysPd4Balant) 2 a speed of

443.4 m/s. What constant force is exerted on the bullet while it is traveling down the 0.7 m length of the barrel of the rifle? Answer in units of N.
Physics
1 answer:
MA_775_DIABLO [31]3 years ago
8 0

Answer:

1109.41 N

Explanation:

v_{o} = initial velocity of the bullet = 0 m/s

v_{f} = final velocity of the bullet as it leaves  = 443.4 m/s

a = acceleration of the bullet

d = length of the barrel of the rifle = 0.7 m

the kinematics equation we can use must include the variables in the above list, hence

{v_{f}}^{2} = {v_{o}}^{2} + 2 a d

{443.4}^{2} = {0}^{2} + 2 (0.7)a

a = 140431.11 ms⁻²

m = mass of the bullet = 7.9 g = 0.0079 kg

Force exerted on the bullet is given as

F = ma

F = (0.0079)(140431.11)

F = 1109.41 N

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yKpoI14uk [10]
  • initial velocity=0m/s=u
  • Acceleration=a=4.8m/s^2
  • Time=t=15s

Final velocity be v

\\ \sf\longmapsto v=u+at

\\ \sf\longmapsto v=0+4.8(15)

\\ \sf\longmapsto v=72m/s

8 0
3 years ago
Air expands adiabatically in a piston–cylinder assembly from an initial state where p1 = 100 lbf/in.2, v1 = 3.704 ft3/lb, and T1
lutik1710 [3]

Answer:

Final temperature is equal to 1291.63°R  

Explanation:

given,

p₁ = 100 lb f/in²,               v₁ = 3.704 ft³/lb,           and T₁ = 1000 °R

p₂ = 30 lb f/in²                 n = 1.4

Δ u = 0.171(T₂ - T₁)

we know for poly tropic process

p vⁿ = constant

p₁ v₁ⁿ = p₂ v₂ⁿ

100 × 3.704¹°⁴ = 30 × v₂¹°⁴

v₂ = 8.753 ft³/lb

work done for poly tropic process

W = \dfrac{p_1v_1-p_2v_2}{n-1}

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W = \dfrac{269.525}{5.40395} Btu/lb

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in the piston cylinder arrangement air is expanding acrobatically

Δ q = Δu + w

Δ u = - w

0.171(T₂ - T₁) = -49.87

0.171(T₁ - T₂) = -49.87

0.171 T₂ = 0.171 × 1000 + 49.87

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5 0
3 years ago
A person 1.8m tall stands 0.75m from a reflecting globe in a garden.
Maru [420]

Answer:

1. The image of the person is 1.41 m, virtual and formed at the back of the surface of the globe.

2. The person's image is 3.38 m tall.

Explanation:

From the given question, object distance, u = 0.75 m, object height = 1.8 m, radius of curvature of the reflecting globe, r = 8 cm = 0.08 m.

f = \frac{r}{2} = \frac{0.08}{2} = 0.04 m

1. The image distance, v, can be determined by applying mirror formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

\frac{1}{0.04} = \frac{1}{0.75} + \frac{1}{v}

\frac{4}{100} - \frac{75}{100} = \frac{1}{v}

\frac{1}{v} = \frac{4 - 75}{100}

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The image of the person is 1.41 m, virtual and formed at the back of the surface of the globe.

2.  \frac{image distance}{object distance} = \frac{image height}{object height}

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v = \frac{2.538}{0.75}

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v = 3.38 m

The person's image is 3.38 m tall.

6 0
3 years ago
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STatiana [176]
\vec F = I (\vec L \times  \vec B)

if currents go in opposite directions, wires repel
3 0
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snow_tiger [21]

Answer: B. CO

Explanation:

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Sodium Chloride NaCl is not a diatomic molecule because is a product of ionization, but it can be diatomic in its gas phase with a polar covalent bond.

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