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timama [110]
3 years ago
12

The crew of a cargo plane wishes to drop a crate of supplies on a target below. To hit the target, when should the crew drop the

crate? Ignore air resistance.
a) Before the plane is directly over the target
b) When the plane is directly over the target
c) After the plane has flown over the target
Physics
2 answers:
lara [203]3 years ago
5 0
To hit the target the crew drop the crate before the plane is directly over the target. It is because <span>because the cargo has forward velocity and therefore before it reaches the ground it travels some distance. The answer is A. Hope it helps. </span>
sergiy2304 [10]3 years ago
5 0

Answer:

a) Before the plane is directly over the target  

Explanation:

When a crate would be dropped from the cargo plane, its initial vertical velocity would be zero (free fall) but it would still have horizontal velocity of the moving plane. Therefore, the falling crate would transverse a parabolic path.

In order to make it hit the target, the crate should be dropped before the plane is directly over the target because of the horizontal component of net velocity will take the crate towards the target.

Thus, option a is correct.

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A line of charge starts at x = +x0 and extends to positive infinity. The linear charge density is λ = λ0x0/x. Determine the elec
kari74 [83]

Explanation:

it is given that, the linear charge density of a charge, \lambda=\dfrac{\lambda_ox_o}{x}

Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :

dE=\dfrac{k\ dq}{x^2}..........(1)

The linear charge density is given by :

\lambda=\dfrac{dq}{dx}

dq=\lambda.dx=\dfrac{\lambda_ox_o}{x}dx

Integrating equation (1) from x = x₀ to x = infinity

E=\int\limits^\infty_{x_o} {\dfrac{k\lambda_ox_o}{x^3}}.dx

E=-\dfrac{k\lambda_ox_o}{2}\dfrac{1}{x^2}|_{x_o}^\infty}

E=\dfrac{k\lambda_o}{2x_o}

Hence, this is the required solution.

5 0
3 years ago
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At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a mass of 0.25 kg, is thrown strai
zhannawk [14.2K]

Answer:

The center of mass of the two-ball system is 7.05 m above ground.

Explanation:

<u>Motion of 0.50 kg ball:</u>

Initial speed, u = 0 m/s

Time = 2 s

Acceleration = 9.81 m/s²

Initial height = 25 m

Substituting in equation s = ut + 0.5 at²

                 s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m

Height above ground = 25 - 19.62 = 5.38 m

<u>Motion of 0.25 kg ball:</u>

Initial speed, u = 15 m/s

Time = 2 s

Acceleration = -9.81 m/s²

Substituting in equation s = ut + 0.5 at²

                 s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m

Height above ground = 10.38 m

We have equation for center of gravity

          \bar{x}=\frac{m_1x_1+m_2x_2}{m_1+m_2}

          m₁ = 0.50 kg

          x₁ = 5.38 m        

          m₂ = 0.25 kg

          x₂ = 10.38 m    

Substituting

         \bar{x}=\frac{0.50\times 5.38+0.25\times 10.38}{0.50+0.25}=7.05m

The center of mass of the two-ball system is 7.05 m above ground.  

8 0
3 years ago
Hey, earn points here
navik [9.2K]

Answer:yayyyy

Explanation:

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Two parallel wires carry a current I in the same direction. Midway between these wires is a third wire, also parallel to the oth
Luda [366]

Answer:

Force is repulsive hence direction of force is away from wire

Explanation:

The first thing will be to draw a figure showing the condition,

Lets takeI attractive force as +ve and repulsive force as - ve and thereafter calculating net force on outer left wire due to other wires, net force comes out to be - ve which tells us that force is repulsive, hence direction of force is away from wire as shown in figure in the attachment.

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75 percent off of water and please water the light water and water water and then go back and please water pollution please 880m
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