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Aleksandr-060686 [28]
3 years ago
6

Three cannons are located at the top of a cliff above a level plain. Cannon A is aimed at an angle of 25° above the horizontal a

nd shoots cannonballs with a speed of 85 m/s. Cannon B is aimed directly horizontally and shoots with a speed of 100 m/s. Cannon C is aimed 25° below the horizontal and shoots with a speed of 75 m/s. Which of the cannonballs will hit the ground with the greatest speed? The three cannonballs are not necessarily the same mass.
Physics
1 answer:
vodka [1.7K]3 years ago
3 0

Answer:

The canon B hits the ground fast.

Explanation:

Given that,

Speed of cannon A = 85 m/s

Speed of cannon B= 100 m/s

Speed of cannon C = 75 m/s

We need to calculate the cannonballs will hit the ground with the greatest speed

Using conservation of energy

The final kinetic energy of canon depends on initial kinetic energy and potential energy.

The  final velocity depends upon initial velocity and initial height.

So,  the initial velocity of canon B is high.

Hence, The canon B hits the ground fast.

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A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
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Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

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E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

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b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

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By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

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E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

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E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

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