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3 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]3 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

<u>The recoil velocity of the gun is 8.33 m/s</u>

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TiliK225 [7]

Answer: 757m/s

Explanation:

Given the following :

Mole of neon gas = 1.00 mol

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Plugging our values

0.5(0.0202 × v^2) = 3/2 (1 × 8.31 × 465)

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3 years ago

A uniform electric field with a magnitude of 5750 N/C points in the positive x direction. Find the change in electric potential

castortr0y [4]

Explanation:

Given that,

Electric field = 5750 N/C

Charge $q=+10.5\times10^{-6}\ C$

Distance = 5.50 cm

(a). When the charge is moved in the positive x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot d$

$\Delta U=-q(E\cdot d)$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times5.50\times10^{-2}$

$\Delta U=-3.32\times10^{-3}\ J$

The change in electric potential energy is $-3.32\times10^{-3}\ J$

(b). When the charge is moved in the negative x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot (-d)$

$\Delta U=-q(E\cdot (-d))$

Put the value into the formula

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The change in electric potential energy is $3.32\times10^{-3}\ J$

Hence, This is the required solution.

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Lelechka [254]

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Given:

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Vladimir79 [104]

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Mice21 [21]

Answer:

Explanation:

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3 years ago