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timama [110]
2 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]2 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]2 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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An arrow is shot at an angle of 35° and a velocity of 50 m/s. How long does it take to return to its original starting height?
Ostrovityanka [42]

Answer:

4.02 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 35°

Initial velocity (u) = 50 m/s

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Time of flight (T) =?

The time of flight of the arrow can be obtained as follow:

T = 2uSineθ / g

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T = 4.02 s

Therefore, the time taken for the arrow to return is 4.02 s

4 0
3 years ago
F1=160N east and F2=50N west, if the acceleration of the object is 3.5 m/s2 what is the mass of the object?
Svetlanka [38]
The mass of the object is the answer of your question

6 0
3 years ago
The efficiency of a squeaky pulley system is 73 percent. The pulleys are usedto raise a mass to a certain height. What force is
larisa86 [58]

The efficiency of the machine is defined as

\eta = \frac{W_{out}}{W_{in}}

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Work out is the work output and Work in is the work input

To find the Work in we have then

W_{in} = \frac{W_{out}}{\eta}

W_{in} = \frac{mgh}{\eta}

Replacing with our values

W_{in} = \frac{(58)(9.8)(3)}{73\%}

W_{in} = 2335.89J

The work done by the applied force is

W = Fd

Here,

F = Force

d = Distnace

Rearranging to find F,

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