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Tpy6a [65]
2 years ago
6

. A girl runs and jumps horizontally off a platform 10m above a pool with a speed of 4.0m/s. As soon as she leaves the platform,

she starts flipping, spinning with a constant angular acceleration of 15.0rad/s2. How many revolutions does she make before hitting the water? (Note that maintaining constant angular acceleration in the air is not realistic, but let’s do it here anyway.)\
Physics
1 answer:
faust18 [17]2 years ago
3 0

Answer:

2.39 revolutions

Explanation:

As she jumps off the platform horizontally at a speed of 10m/s, the gravity is the only thing that affects her motion vertically. Let g = 10m/s2, the time it takes for her to fall 10m to water is

h = gt^2/2

10 = 10t^2/2

t^2 = 2

t = \sqrt{2} = 1.414 s

Knowing the time it takes to fall to the pool, we calculate the angular distance that she would make at a constant acceleration of 15 rad/s2:

\theta = \alpha t^2/2

\theta = 15 * 2/2 = 15 rad

As each revolution is 2π, the total number of revolution that she could make is: 15 / 2π = 2.39 rev

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The total work is

         (mass of the elevator, kg) x (9.8 m/s²) x (9.0 m)           Joules .
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PLEASE HELPPP 100 POINTS HURRY !!!!Which diagram best illustrates the magnetic field of a bar magnet? A bar magnet with a north
Serggg [28]

I think this is right I hope this is right for you

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2 years ago
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Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀)  we see that for the same t v₁> v₂

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You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t²- 2 t to + to²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to.  The rock y has not left and the distance increases

For t> = to.  the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t   v₁> v₂

3 0
2 years ago
A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the
Fofino [41]

Answer:

2 x 10⁻³ volts

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θ = Angle of the magnetic field with the area vector = 0

E = emf induced in the loop

Induced emf is given as

E = B \frac{dA}{dt}

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

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7 0
2 years ago
A 0.305 kg book rests at an angle against one side of a bookshelf. The magnitude and direction of the total force exerted on the
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Answer

given,

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so, from the diagram attached  below

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cos {\theta_b} = \dfrac{1.52 \times sin {31^0}}{1.686}

cos {\theta_b} = 0.464

θ = 62.35°

5 0
2 years ago
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