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2 years ago

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If he leaves the ramp with a speed of 35.0 m/s and has a speed of 33.0 m/s at the top of his trajectory, determine his maximum h

eight (h) (in m) above the end of the ramp. Ignore friction and air resistance.

1 answer:

nadezda [96]2 years ago

7 0

Answer:

H = 6.93 m

Explanation:

given data

velocity v = 35 m/s

horizontal component Vx = 33 m/s

solution

we get here maximum height so first we get vertical component here that is express as

Vy = $\sqrt{v^2- Vx^2}$ .........................1

put here value

Vy = $\sqrt{35^2- 33^2}$

Vy = 11.66 m/s

and

now we get height

H = $\frac{Vy^2}{2g}$ .............................2

put here value

H = $\frac{11.66^2}{2\times 9.8}$

H = 6.93 m

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A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a

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Frequency = 60 Hz

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$\omega=2\pi f$

Put the value into the formula

$\omega=2\times\pi\times60$

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Using formula of magnetic filed

$B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}$

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V = voltage

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3 years ago