Answer: Force applied by trampoline = 778.5 N
<em>Note: The question is incomplete.</em>
<em>The complete question is : What force does a trampoline have to apply to a 45.0 kg gymnast to accelerate her straight up at 7.50 m/s^2? note that the answer is independent of the velocity of the gymnast. She can be moving either up or down or be stationary.
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Explanation:
The total required the trampoline by the trampoline = net force accelerating the gymnast upwards + force of gravity on her.
= (m * a) + (m * g)
= m ( a + g)
= 45 kg ( 7.50 * 9.80) m/s²
Force applied by trampoline = 778.5 N
Answer:
My scenario would be A Car vs. a guard rail on a road. You have a car that is coming down a Highway at a speed of 43 Mph Miles per hour (69.2018 Kmh)
And it hits a steel guardrail and the car smashes in at the front and the guardrail is only bent while the car has the bumper and the hood along with the headlights and windshield along with the passenger side window break.
Explanation:
This is caused by so much force reacting from one object to another but also depends on molecular density.
It took so long because at the time there was no way for people to study the behavior formally. im not sure what helped it get recognized but i know wihelm wundt helped get it recongnized.
sorry i couldnt be much help
Answer:
Explanation:
Area A of the coil = .1 x .1 = .01 m²
no of turns n = 5
magnetic field B = .5 t²
Flux Φ perpendicular to plane passing through it.= nBA sin30
rate of change of flux
dΦ/dt = nAdBsin30 / dt
= nA d/dt (.5t²x .5 )
= nA x 2 x .25 x t
At t = 4s
dΦ/dt = nA x 2
= 5x .01 x 2
= .1
current = induced emf / resistance
= .1 / 4
= .025 A
= 25 mA.
Answer:
Magnetic field, B = 0.23 T
Explanation:
Given that,
Length of the copper rod, L = 0.49 m
Mass of the copper rod, m = 0.15 kg
Current in rod, I = 13 A (in +ve y direction)
When the rod is placed in magnetic field, the magnetic force is balanced by its weight such that :

So, the magnitude of the minimum magnetic field needed to levitate the rod is 0.23 T.