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3 years ago

5.0 kg, with a bullet of mass 0.1 kg. The target was mounted on

Physics

1 answer:

kari74 [83]3 years ago

8 0

Answer:

306 m/s

Explanation:

Law of conservation of momentum

m1v1 + m2v2 = (m1+m2)vf

m1 is the bullet's mass so it is 0.1 kg

v1 is what we're trying to solve

m2 is the target's mass so it is 5.0 kg

v2 is the targets velocity, and since it was stationary, its velocity is zero

vf is the velocity after the target is struck by the bullet, so it is 6.0 m/s

plugging in, we get

(0.1 kg)(v1) + (5.0 kg)(0 m/s) = (0.1 kg + 5.0 kg)(6.0 m/s)

(0.1)(v1) + 0 = 30.6

(0.1)(v1) = 30.6

v1 = 306 m/s

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Put the value into the formula

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3 years ago

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3 years ago

Read 2 more answers

A torpedo is to be designed to be 3 m long with a diameter of 0.5 m. Treating the torpedo as a cylinder, it is to be made to hav

rodikova [14]

<h3>Answer:</h3>

The velocity that will be needed for the model and prototype to be similar is 108.97m/s

Explanation:

length of Torpedo = 3m

diameter, $d_{1} = 0.5m$

velocity of sea water, $v_{1}$ = 10m/s

dynamic viscosity of sea water, η $_{1}$ = 0.00097 Ns/m²

density of sea water, ρ $_{1}$ = 1023 kg/m³

Scale model = 1:15

$\frac{d_{1} }{d_{2} }$ = $\frac{1}{15}$

Cross multiplying: d $_{2}$ = $15d_{1} }$ = 15 ×0.5 = 7.5m

Let:

velocity of air, $v_{2}$

viscosity of air, η $_{2}$ = 0.000186Ns/m²

density of air, ρ $_{2}$ = 1.2 kg/m³

For the model and the prototype groups to be equal, Non-dimensional groups should be equal.

Reynold's number: (ρ $_{2}$ × $v_{2}$ ×d $_{2}$ )/η $_{2}$ = (ρ $_{1}$ × $v_{1}$ ×d $_{1}$ )/η $_{1}$

$v_{2}$ = η $_{2}$ /(ρ $_{2}$ ×d $_{2}$ ) × (ρ $_{1}$ × $v_{1}$ ×d $_{1}$ )/η $_{1}$

$v_{2}$ = $\frac{0.000186}{1.2* 7.5}$ × $\frac{1023 *10*0.5}{0.00097}$ , note: * means multiplication

$v_{2}$ = 108.97m/s

velocity that will be needed for the model and prototype to be similar = 108.97m/s

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3 years ago