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defon
2 years ago
13

When a cannon fires a cannonball we observe Newton's third law. Which is the reaction force? A) The cannon moving forward B) The

cannonball moving backwards C) The cannon moving backwards D) The cannonball moving forward
Physics
2 answers:
laiz [17]2 years ago
7 0

Answer:

I

will

only

explain

Explanation:

The cannon is fired when an explosive charge is detonated, causing a sudden and immense increase in pressure. Is it not this pressure that causes the rapid acceleration of the cannonball and the recoil of the cannon, not an action–reaction between the cannon and the cannonball?

For purposes of this model, we can consider the expanding gas from the explosion to be part of the cannon, or as an intervening object between the cannon and the ball. So the gasses exert a force on the ball. The ball exerts a force back on the gasses. This is transferred to the cannon.

You could also imagine or build a (toy) "cannon" with a spring mechanism to propel the ball, rather than an explosion. You'd see very similar results.

In any case, the deeper point, which you will soon learn, is that momentum is a conserved quantity. Regardless of what mechanism applies the force on the ball and the cannon, after the ball is flying free the cannon must end up with as much backwards momentum as the ball has forward momentum.

If no cannonball is present when the charge is detonated, then the pressure dissipates much more quickly and the recoil is smaller but still present

Because air and exhaust gasses from the explosion are expelled from the cannon. These gasses have mass and carry momentum, therefore they exert a reaction force on the cannon just as a ball does.

HOPE IT HELPS

<h2> </h2>

<h2><em>it's </em><em>hard </em><em>but </em><em>not </em><em>for </em><em>me</em></h2>

<h2><em>mark </em><em>me </em><em>in </em><em>brainliest </em><em>answers </em><em>please </em><em>please </em><em>please </em></h2>

<h2 />
scoundrel [369]2 years ago
6 0

Answer:

I cant answer too complicated sorry

Explanation:

im sorry for not answering

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Answer:

1. a

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3 0
3 years ago
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin
lyudmila [28]
Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
W= \int\limits^{0.540m}_{0} {F} \, dx =  \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =
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By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
W=16000(0.540m)+10000  \frac{(0.540m)^2}{2} - 26000  \frac{(0.540m)^3}{3}  =
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part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
W=16000(0.95m)+10000  \frac{(0.95m)^2}{2} - 26000  \frac{(0.95m)^3}{3}  =
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5 0
3 years ago
Combine these three velocity vectors into a resultant: 3.0 m/s north, 4.0 m/s east 1.0 m/s west. Identify the resultant vector
Slav-nsk [51]

Answer:

The resultant vector is 1 m/s

Explanation:

The resultant vector is 1 m/s west based on triangle law of vector addition, when two sides of a triangle is represented by two vectors, the resultant vector is the third side of the triangle.

5 0
3 years ago
A garden hose having with an internal diameter of 1.1 cm is connected to a (stationary) lawn sprinkler that consists merely of a
trapecia [35]

Answer:

Water leaves the sprinkler at a speed of 2.322 m/sec

Explanation:

We have given internal diameter of the garden d_1=1.1cm

Speed of water in the hose is v_1=0.95m/sec

Number of holes n = 22

Diameter of each holes d_2=15cm

According to continuity equation A_1v_1=A_2v_2

d_1^2\times v_1=22\times d_2^2v_2

1.1^2\times 0.95=22\times 0.15^2\times v_2

v_2=2.322m/sec

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yarga [219]

Answer: the answer is a

Explanation:

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