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ruslelena [56]
2 years ago
9

A 15kg hoop with a radius of 3 m is rolling at on a level surface when it reaches a 25 degree incline. If it reaches a height of

4 m, what was the hoop's initial velocity?
Physics
1 answer:
dimulka [17.4K]2 years ago
4 0

The initial velocity of the hoop is determined as 8.854 m/s.

<h3>Conservation of energy</h3>

The initial velocity of the hoop can be determined from the principle of conservation of energy.

Final potential energy = Initial kinetic energy

P.E = K.E

mgh = ¹/₂mv²

gh = ¹/₂v²

2gh = v²

√2gh = v

√(2 x 9.8 x 4) = v

8.854 m/s = v

Thus, the initial velocity of the hoop is determined as 8.854 m/s.

Learn more about initial velocity here: brainly.com/question/19365526

#SPJ1

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neonofarm [45]

Answer:

v = 3.04 m/s

Explanation:

given,

mass of the block, M = 6.6 Kg

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we now

Work done is equal to change in Kinetic energy.

Work done = Force x displacement

W = Δ K E

Δ K E = Force x displacement

\dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2= F .s

\dfrac{1}{2}\times 6.6 \times v^2 - 0= 12.2\times 2.5

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4 0
2 years ago
To practice Problem-Solving Strategy 30.1: Inductors in Circuits. A circuit has a 1 V battery connected in series with a switch.
ki77a [65]

Answer:

0.0133A

Explanation:

Since we have two sections, for the Inductor region there would be a current i_1. In the case of resistance 2, it will cross a current i_2

Defined this we proceed to obtain our equations,

For i_1,

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

For i_2,

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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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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}}})

i= 0.0133A

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3 0
3 years ago
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dsp73

Answer:

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Katen [24]
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s2008m [1.1K]

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