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andre [41]
3 years ago
13

A battleship simultaneously fires two shells toward two identical enemy ships. One shell hits ship A, which is close by, and the

other hits ship B, which is farther away. The two shells are fired at the same speed. Assume that air resistance is negligible and that the magnitude of the acceleration due to gravity is g. What shape is the trajectory (graph of y vs. x) of the shells?a. straight lineb. parabolac. hyperbolad. The shape cannot be determined.
Physics
1 answer:
horsena [70]3 years ago
8 0

Answer:

Parabola.

Explanation:

No matter how far away are the enemy ships, if they are identical or not, or how much is the initial velocity, if the shells are subject to the force of gravity then the shape of their trajectories will follow the same form as with everything else: a parabola.

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Explanation:

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A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between
Oduvanchick [21]

Answer:

Net electric field, E_{net}=91406.24\ N/C

Explanation:

Given that,

Charge 1, q_1=7.5\ nC=7.5\times 10^{-9}\ C

Charge 2, q_2=-2.9\ nC=-2.9\times 10^{-9}\ C

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

E_1=\dfrac{kq_1}{r^2}

E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}

E_1=65917.96\ N/C

Electric field due to charge 2 is given by :

E_2=\dfrac{kq_2}{r^2}

E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}

E_2=25488.28\ N/C

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

E_{net}=E_1+E_2

E_{net}=65917.96+25488.28

E_{net}=91406.24\ N/C

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

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The correct options are:

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