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2 years ago

A shotgun of mass 3kg fires a bullet of 0.1kg at a velocity of 250m/s. What is the recoil velocity of the gun

Physics

1 answer:

alisha [4.7K]2 years ago

6 0

Answer:

8.33 m/s

Explanation:

The formula to apply is that of conservation of linear momentum taking that the this case is a single fire non reciprocating action.

Thus,

m₁v₁ = m₂v₂ where;

m₁ = mass of the shotgun = 3kg

v₁ = recoil velocity of the gun

m₂ = mass of bullet

v₂ = velocity of bullet

Substitute values in the equation as;

m₁v₁ = m₂v₂

3*v₁ =0.1*250

3v₁ =25

v₁ =25/3 = 8.33 m/s

The recoil velocity of the gun is 8.33 m/s

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2 years ago

Glycerin (C3H8O3) is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 154 g of glycerin to 316 mL o

garri49 [273]

Answer:

$P_{sol}=50.4\ mm.Hg$

Explanation:

According to given:

molecular mass of glycerin, $M_g=3\times 12+8+3\times 16=92\ g.mol^{-1}$

molecular mass of water, $M_w=2+16=18\ g.mol^{-1}$

∵Density of water is $0.992\ g.cm^{-3}= 0.992\ g.mL^{-1}$

∴mass of water in 316 mL, $m_w=316\times 0.992=313.5 g$

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pressure of mixture, $P_x=55.32\ torr= 55.32\ mm.Hg$

temperature of mixture, $T_x=40^{\circ}C$

Upon the formation of solution the vapour pressure will be reduced since we have one component of solution as non-volatile.

moles of water in the given quantity:

$n_w=\frac{m_w}{M_w}$

$n_w=\frac{313.5}{18}$

$n_w=17.42 moles$

moles of glycerin in the given quantity:

$n_g=\frac{m_g}{M_g}$

$n_g=\frac{154}{92}$

$n_g=1.674 moles$

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$X_w=\frac{n_w}{n_w+n_g}$

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Since glycerin is non-volatile in nature so the vapor pressure of the resulting solution will be due to water only.

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3 years ago

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3 years ago

How do I solve this arithmetic sequence?

Serjik [45]

Answer:

The 16ᵗʰ term of this sequence is 82

Step-by-step explanation:

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First Term = a₁ = 9

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Now, For 16ᵗʰ term, n = 16

aₙ = a + (n - 1)d

a₁₆ = 7 + (16 - 1) × 2

a₁₆ = 7 + 15 × 5

a₁₆ = 7 + 75

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Thus, The 16ᵗʰ term of this sequence is 82

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2 years ago