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2 years ago

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How is shooting a shotgun related to newtons third law ? why does a rifle have less “kick” then a shotgun ?

1 answer:

Mazyrski [523]2 years ago

5 0

Newton’s third law is a force is a push or pull that acts upon an object as a result of another object.. A shot gun when fired pulls back when a force (you pulling the trigger) is acted upon it. This is how it relates. A rifle has less kick than a shot gun because the rifle is smaller and has less of a force than a shot gun.

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A ball is thrown upwards at an unknown speed. in a time of

alexandr402 [8]

Answer:

a) Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Explanation:

a) We have equation of motion $s=ut+\frac{1}{2} at^2$ , where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8 $m/s^2$

Substituting

$6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s$

Initial speed of the ball = 14.45 m/s

b) We have equation of motion $v^2=u^2+2as$ , where v is the final velocity

s = 6 m, u = 14.45 m/s, a = -9.8 $m/s^2$

Substituting

$v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s$

So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion $v^2=u^2+2as$ , where v is the final velocity

u = 14.45 m/s, v =0 , a = -9.8 $m/s^2$

Substituting

$0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m$

Maximum height reached = 10.65 m

8 0

2 years ago

How much heat (in calories) are needed to raise the temperature of 60 g of water

laila [671]

Answer:

Well, each ml of water requires one calorie to go up 1 degree Celsius, so this liter of water takes 1000 calories to go up 1 degree Celsius.

Explanation:

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2 years ago

What velocity must a 2.25 kg croquet mallet have in order to have the same momentum as 1.25 kg ball that has a momentum 6.25 kg•

Ksenya-84 [330]

abcdefghijklmnopqrstuvwxyz

3 0

2 years ago

A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com

bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature $=20^{\circ}C$

Final Temperature $=80^{\circ}C$

mass flow rate of cold fluid $\dot{m_c}=1.2 kg/s$

Initial Geothermal water temperature $T_h_i=160^{\circ}C$

Let final Temperature be T

mass flow rate of geothermal water $\dot{m_h}=2 kg/s$

diameter of inner wall $d_i=1.5 cm$

$U_{overall}=640 W/m^2K$

specific heat of water $c=4.18 kJ/kg-K$

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

$\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)$

$2\times (160-T)=1.2\times (80-20)$

$160-T=36$

$T=124^{\circ}C$

As heat exchanger is counter flow therefore

$\Delta T_1=160-80=80^{\circ}C$

$\Delta T_2=124-20=104^{\circ}C$

$LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}$

$LMTD=\frac{80-104}{\ln \frac{80}{104}}$

$LMTD=91.49^{\circ}C$

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

$\dot{m_c}c(80-20)=U\cdot A\cdot$ (LMTD)

$A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2$

$A=\pi DL=5.144$

$L=\frac{5.144}{\pi \times 0.015}$

$L=109.16 m$

6 0

2 years ago

1.20 Newton force is working on a 250 gram object. What is the acceleration?

Leokris [45]

Answer:

The answer is B

Explanation:

250g = 0.25kg

F = m × a

a = F/m

= 1.2/0.25

= 4.8m/s²

7 0

2 years ago