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UkoKoshka [18]
2 years ago
6

In a skate park, you are trying to determine how to get the most speed at the bottom of the ramp. If the ramp is 4.5 m high, and

you start at the top of the ramp with an initial velocity of 1.5m/s, how fast will you be rolling at the bottom of the ramp, given there is no friction. Use conservation of energy to solve.
Physics
1 answer:
Oksanka [162]2 years ago
7 0

Answer:

If you have an initial velocity of 1.5m/s and the ramp height is 4.5, and an extreme fart is initiated at the bottom of the ramp exerting an didditional upward force, and the smell of the fart causes you to begin flailing, and everyone running from you causes an additional wind force, then you would be rolling at a speed of 3 m/s given there is no friction. :)

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Answer:

Answer:

4 ms

Explanation:

initial velocity, u = 75 m/s

final velocity, v = 0

distance, s = 15 cm = 0.15 m

Let the acceleration is a and the time taken is t.

Use third equation of motion

v² = u² + 2 a s

0 = 75 x 75 - 2 a x 0.15

a = - 18750 m/s^2

Use first equation of motion

v = u + at

0 = 75 - 18750 x t

t = 4 x 10^-3 s

t = 4 ms

thus, the time taken is 4 ms.

Explanation:

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To practice Problem-Solving Strategy 30.1: Inductors in Circuits. A circuit has a 1 V battery connected in series with a switch.
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Answer:

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Explanation:

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Defined this we proceed to obtain our equations,

For i_1,

\frac{di_1}{dt}+i_1R_1 = V

I_1 = \frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})

For i_2,

I_2R_2 =V

I_2 = \frac{V}{R_2}

The current in the entire battery is equivalent to,

i_t = I_1+I_2

i_t = \frac{V}{R_2}+\frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})

Our values are,

V=1V

R_1 = 95\Omega

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R_2 =360\Omega

Replacing in the current for t= 0.4m/s

i=\frac{1}{360}+\frac{1}{95}(1-e^{-\frac{95*0.4}{1.5*10^{-2}}})

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3 years ago
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Answer:

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Explanation:

formula is

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F = 6.49 x 10^-8 Newtons

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