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JulijaS [17]
2 years ago
5

The recoil of a cannon is probably familiar to anyone who has watched pirate movies. This is a classic problem in momentum conse

rvation. Consider a wheeled, 500 kg cannon firing a 2 kg cannonball horizontally from a ship. The ball leaves the cannon traveling at 200 m/s. At what speed does the cannon recoil as a result?
Physics
1 answer:
babunello [35]2 years ago
8 0

Answer:

0.8 m/s

Explanation:

From the question,

MV = mv............................. Equation 1

Where M = mass of the cannon, m = mass of the cannonball, V = velocity of the cannon, v = velocity of the cannon ball.

make V the subject of the equation

V = mv/M........................ Equation 2

Given: m = 2 kg, M = 500 kg, v = 200 m/s

Substitute these value into equation 2

V = (2×200)/500

V = 400/500

V = 0.8 m/s

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Below is the solution:

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<span>12*0.35*3800*(t-5)=6.5*4200*(27-t) </span>
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4 0
3 years ago
The velocity of a 1.3 kg remote-controlled car is plotted on the graph. The work of segment A is J.
tamaranim1 [39]

Answer: 585 J

Explanation:

We can calculate the work done during segment A by using the work-energy theorem, which states that the work done is equal to the gain in kinetic energy of the object:

W=K_f -K_i

where Kf is the final kinetic energy and Ki the initial kinetic energy. The initial kinetic energy is zero (because the initial velocity is 0), while the final kinetic energy is

K_f =\frac{1}{2}mv^2

The mass is m=1.3 kg, while the final velocity is v=30 m/s, so the work done is:

W=K_f = \frac{1}{2}(1.3 kg)(30 m/s)^2=585 J

5 0
3 years ago
Read 2 more answers
Find the average speed of a walk to school and back if you took 12 minutes to walk there and 18 minutes to get back
nikdorinn [45]

Answer:

12+ 18 divide by 2 is the average minutes

7 0
2 years ago
A crate is sliding on the floor. If there is a total force acting on the crate in the same direction as it is sliding, the crate
Leto [7]

But we do not know whether the force is pushing or pulling (the same direction (both forces are parallel) but: .........[ ]<-F-- or .......[ ]--F-->). I suppose the correct answer is B

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3 years ago
A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou
Hoochie [10]

Answer:

e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.

Explanation:

This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.

The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.

<u>Mathematical expression for the Newton's second law of motion is given as:</u>

F=\frac{dp}{dt} ............................................(1)

where:

dp = change in momentum

dt = time taken to change the momentum

We know, momentum:

p=m.v

Now, equation (1) becomes:

F=\frac{d(m.v)}{dt}

<em>∵mass is constant at speeds v << c (speed of light)</em>

\therefore F=m.\frac{dv}{dt}

and, \frac{dv}{dt} =a

where: a = acceleration

\Rightarrow F=m.a

also

F\propto \frac{1}{dt}

so, more the time, lesser the force.

<em>& </em><u><em>Impulse:</em></u>

I=F.dt

I=m.a.dt

I=m.\frac{dv}{dt}.dt

I=m.dv=dp

∵Initial velocity and final velocity(=0), of a certain mass is same irrespective of the stopping method.

So, the impulse in both the cases will be same.

4 0
3 years ago
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