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Tems11 [23]
3 years ago
12

Assuming the dread pirate roberts never misses, how far from the end of the cannon is the ship that you are trying to hit (negle

ct dimensions of cannon)? answer in units of m.
Physics
1 answer:
KengaRu [80]3 years ago
8 0
<span>First let's find the acceleration required in the barrel to speed the ball up from 0 to 83 m/s in a distance of 2.17 m. We know the force the cannon exerts on the cannonball is 20000 N; if we can find this acceleration then we can use F = ma to find the mass. We can find the acceleration using one of the kinematic equations of motion. We have: u = initial speed = 0 m/s v = final speed = v0 = 83 m/s d = distance = 2.17 m a = acceleration = ? v² = u² + 2ad. Since u = 0, this reduces to v² = 2ad and rearranges to a = v²/2d = 83²/2*2.17 = 83²/4.34 = 1587.327 m/s². Now F = ma, so m = F/a = (20000N)/(1587.327 m/s²) = 12.6 kg. For part 2, use the Range Equation: If R is the horizontal distance the cannonball travels, v = v0 = the initial velocity = 83 m/s g = acceleration due to gravity - 9.8 m/s² x the launch angle relative to the horizontal, then R = (v²sin(2x))/g. So R = (83²sin(2*37))/9.8 = (6889sin74)/9.8 = 676 m. So the target ship is 676 m away.</span>
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RSB [31]

The velocities at time <em>t</em> are

• Horizontal:

<em>v</em> = (30.0 m/s) cos(20.0º)

• Vertical:

<em>v</em> = (30.0 m/s) sin(20.0º) - <em>g</em> <em>t</em>

(where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity)

If you only want the <u>initial</u> velocities, they are

• Horizontal:

<em>v</em> = (30.0 m/s) cos(20.0º) ≈ 28.2 m/s

• Vertical:

<em>v</em> = (30.0 m/s) sin(20.0º) ≈ 10.3 m/s

(just set <em>t</em> = 0)

As far as starting equations go, you can derive everything from the definition for average acceleration:

<em>a</em> = ∆<em>v</em> / ∆<em>t</em> = (final <em>v</em> - initial <em>v</em>) / <em>t</em>

→   <em>v</em> = <em>u</em> + <em>a</em> <em>t</em>

(here, <em>u</em> stands in for "initial <em>v</em>" and <em>v</em> is simply velocity at time <em>t</em> )

There is no acceleration in the horizontal direction, while the ball is essentially in free-fall in the vertical direction.

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3 years ago
Two different charges, q1 and q2, are placed at two different locations, one charge at each location. The locations have the sam
Mars2501 [29]

Answer:

No

Explanation:

Electric potential is the work done to bring a unit of charge (1 C) from infinity to a point inside an electric field.

Electric potential energy of a charge q is the energy required to keep it in an electric potential V. Electric potential energy is given by,

U = qV

Hence even if the two charges are on an equipotential surface (surface where the potential is the same at all points), the potenial energy will be different if the magnitude or nature of the charges are different.

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The behaviour that has just been displayed by the girl is an example of a learned behavior.

<h3>What is learned behaviour?</h3>

The term learned behaviour has to do with those behaviour that the individual acquires by practise and are not innate in the organism. The ability to drink a large amount of water without breathing is not inate in man.

Hence, the behaviour that has just been displayed by the girl is an example of a learned behavior.

Learn more about learned behavior:brainly.com/question/347230

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Differences between single &amp; double displacement reaction​
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