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cupoosta [38]
2 years ago
12

A 3.00-g bullet has a muzzle velocity of 230 m/s when fired by a rifle with a weight of 25.0 N. (a) Determine the recoil speed o

f the rifle. 0.27 Correct: Your answer is correct. m/s (b) If a marksman with a weight of 650 N holds the rifle firmly against his shoulder, determine the recoil speed of the shooter and rifle. Incorrect: Your answer is incorrect.
Physics
1 answer:
goblinko [34]2 years ago
7 0

Answer:

0.27 m/s and 0.01 m/s

Explanation:

Given that

Mass of the bullet, m = 3 g = 0.003 kg

Velocity of the bullet, v = 230 m/s

Weight of the rifle, w = 25 N

From this, we know that the mass of the rifle, M = 25/9.8 = 2.55 kg

Using the law of conservation of momentum, we know that

Momentum of the gun = momentum of the bullet

M.g * V.g = m.b * v.b

2.55 * V.g = 0.003 * 230

2.55V.g = 0.69

V.g = 0.69 / 2.55

V.g = 0.27 m/s

If weight of the shooter = 650, then his mass = 650/9.8 = 66.33 kg

The recoil speed of the shooter is

0.69 / 66.33 = 0.010 m/s

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2 years ago
A gry is an old English measure for length, defined as 1/10 of a line, where line is another old English measure for length, def
Valentin [98]

Answer:

0.2448 point²

Explanation:

1 gry = 1/10 line

1 line = 1/12 inch

=> 1 gry in inches = 1/10 * 1/12 = 1/120 inch

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1 point = 1/72 inch

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Therefore,

120 gry = 72 points

=> 1 gry = 3/5 point

Therefore,

1 gry² = (3/5)² point²

1 gry² = 9/25 point²

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0.68 gry² = 0.68 * 9/25 point²

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3 years ago
You and your friend are pushing a box at the same constant speed. You pushed 20 feet in 10 minutes, your friend was pushing for
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Answer:

60 N.C

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2 years ago
A kayakeris paddling 2.50 m/s at an angle of 45° (northeast) and the current is moving 1.25 m/s at an angle of 315° (southeast).
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The kayaker has velocity vector

<em>v</em> = (2.50 m/s) (cos(45º) <em>i</em> + sin(45º) <em>j</em> )

<em>v</em> ≈ (1.77 m/s) (<em>i</em> + <em>j</em> )

and the current has velocity vector

<em>w</em> = (1.25 m/s) (cos(315º) <em>i</em> + sin(315º) <em>j</em> )

<em>w</em> ≈ (0.884 m/s) (<em>i</em> - <em>j</em> )

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<em>v</em> + <em>w</em> ≈ (2.65 m/s) <em>i</em> + (0.884 m/s) <em>j</em>

That is, the kayaker has a velocity of about ||<em>v</em> + <em>w</em>|| ≈ 2.80 m/s in a direction <em>θ</em> such that

tan(<em>θ</em>) = (0.884 m/s) / (2.65 m/s)   →   <em>θ</em> ≈ 18.4º

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