Answer:
The effective spring constant of the firing mechanism is 1808N/m.
Explanation:
First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

(This is correct because the horizontal motion has acceleration zero). Then:

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

Then, plugging in the given values, we obtain:

Finally, the effective spring constant of the firing mechanism is 1808N/m.
Answer:
∴ [T]=[WF−1V−1]
Hope this answer is right!!
<span>An automobile with a mass of 1450 kg is parked on a moving flatbed railcar; the flatbed is 1.5 m above the ground. The railcar has a mass of 38,500 kg and is moving to the right at a constant speed of 8.7 m/s on a frictionless rail...
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Answer: F
Out of the page.
Explanation:
For an electron with a charge of -e, the magnitude of the force on it is F = BeV
Where
F = force on the electron
e = charge ( electrons )
V = velocity
B = magnetic field
F is the force acting on all the electrons in a wire which gives rise to the F = BIL
Where
I = current
L = length of the wire
The force F is always at the right angle to the particle's velocity and its direction can be found using the left hand rule.
When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed out of the page.
2.0y i think this is guess but if it right then thats good