Answer
D. move a small magnet back and forth within a section of the coiled wire.
Explanation:
i put that for the test and i got it right
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
Acceleration due to gravity (g) = 10 m/s²
Time of flight (T) =?
The time of flight of the arrow can be obtained as follow:
T = 2uSineθ / g
T = 2 × 35 × Sine 35 / 10
T = 70 × 0.5736 / 10
T = 7 × 0.5736
T = 4.02 s
Therefore, the time taken for the arrow to return is 4.02 s
The mass of the object is the answer of your question
The efficiency of the machine is defined as

Here
Work out is the work output and Work in is the work input
To find the Work in we have then


Replacing with our values


The work done by the applied force is
W = Fd
Here,
F = Force
d = Distnace
Rearranging to find F,


F = 129.77N
Therefore the force exerted on the machine after rounding off to two significant figures is 130N