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Debora [2.8K]
3 years ago
6

When a gun is fired at the shooting range, the gun recoils (moves backward). Explain this using the law of conservation of momen

tum
Physics
1 answer:
12345 [234]3 years ago
4 0
The total momentum is unchanged according to the law of conservation of momentum. When the gun is fired, the bullet gains a high velocity forward (positive velocity), and that velocity multiplied by its mass is the momentum the bullet gains. Therefore, the gun must gain a momentum backwards to cancel out that momentum forward, so the gun recoils back with a negative velocity.
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What is stopping distance of a body at 20 m/s is decelerated at 2m/s² to rest?<br>​
AlladinOne [14]

Answer:

s = 100 m

Explanation:

v² = u² + 2as

s = (v² - u²) / 2a

s = (0² - 20²) / 2(-2)

s = 100 m

5 0
3 years ago
Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci
Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W

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h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}

2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

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6 0
3 years ago
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Alja [10]

Answer:

\vec{v} =

Explanation:

given,

location of the ball ⟨7,0,−8⟩

initial velocity of the ball ⟨-11,19,−5⟩

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speed of the ball = ?

using Momentum Principle

change in momentum = Force x time

m \vec{v} - m \vec{u}= \vec{F}\times \Delta t

\vec{v} =\vec{u} + \dfrac{\vec{F}}{m}\times \Delta t

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F_net = F_g = ⟨0 ,-9.8 m , 0⟩

now,

\vec{v} = + \dfrac{}{m}\times (0.4-0)

\vec{v} = +

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Explanation:

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