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loris [4]
2 years ago
6

Describe a theme-park ride that has constant speed but changing velocity. Thank you!

Physics
1 answer:
photoshop1234 [79]2 years ago
8 0
How about a carousel (merry go round).
For any one horse or rider, Speed is constant but direction keeps changing, so velocity does too.
You might be interested in
What are some other examples of scientific knowledge that must have resulted from indirect evidence?
n200080 [17]

Answer:

The atomic model, the structure of the solar system, and the theory of black holes are examples of models created using indirect evidence.

Explanation:

I have no explanation sorry.

5 0
2 years ago
In the winter sport of bobsledding, athletes push their sled along a horizontal ice surface and then hop on the sled as it start
Hitman42 [59]

Answer:

v_f = 16.6 m/s

Explanation:

As we know by force equation that force along the inclined planed due to gravity is given as

F_g = mg sin\theta

so the acceleration due to gravity along the plane is given as

a = \frac{F_g}{m}

now we have

a = g sin\theta

a = (9.81 sin4.0)

a = 0.68 m/s^2

now we know that

v_f^2 - v_i^2 = 2 a d

v_f^2 - 9.2^2 = 2(0.68)(140)

v_f = 16.6 m/s

4 0
2 years ago
A turntable that is initially at rest is set in motion with a constant angular acceleration α. What is the magnitude of the angu
bekas [8.4K]

Explanation:

If the turntable starts from rest and is set in motion with a constant angular acceleration α. Let \omega is the angular velocity of the turntable. We know that the rate of change of angular velocity is called the angular acceleration of an object. Its formula is given by :

\alpha =\dfrac{\omega_f-\omega_i}{t}

\alpha =\dfrac{\omega-0}{t}

\alpha =\dfrac{\omega}{t}

t=\dfrac{\omega}{\alpha }............(1)

Using second equation of kinematics as :

\theta=\omega_i t+\dfrac{1}{2}\alpha t^2

\theta=\dfrac{1}{2}\alpha t^2

Using equation (1) in above equation

\theta=\dfrac{1}{2}\times \dfrac{\omega^2}{\alpha }

In one revolution, \theta=4\pi (in 2 revolutions)

4\pi =\dfrac{1}{2}\times \dfrac{\omega^2}{\alpha }

\omega=\sqrt{8\pi \alpha}

\omega=2\sqrt{2\pi \alpha}

Hence, this is the required solution.

6 0
2 years ago
Read 2 more answers
A bowling ball starts from rest and moves 300 m down a long, downwardly angled track in 22.4 sec.a. What is its speed at the end
Sergio [31]

Answer:

The speed at the end of the track = 27 m/s

The acceleration = 1.2 m/s²

Please find the Δx vs Δt, v vs Δt, a  vs Δt

Explanation:

We have;

x = u·t + 1/2·a·t²

Where;

x = The distance = 300 m

u = The initial velocity = 0 m/s (Ball at rest)

t = The time taken = 22.4 s

Therefore;

300 = 0 + 1/2×a×22.4²

a = 2×300/22.4² = 1.19579 ≈ 1.2 m/s²

v = u + a×t

∴ v = 0 + 1.2 × 22.4 = 26.88 ≈ 27 m/s

Part of the table of values is as follows;

t,                                     x,                                          v

0,                                    0,                                        0

0.4,                                 0.095663,                         0.478316

0.8,                                 0.382653,                          0.956632

1.2,                                   0.860969,                         1.434948

1.6,                                   1.530611,                            1.913264

2,                                     2.39158,                              2.39158

2.4,                                  3.443875,                           2.869896

2.8,                                   4.687497,                          3.348212

3.2,                                   6.122445,                           3.826528

3.6,                                   7.748719,                             4.304844

4 0
2 years ago
Suppose that the mirror described in Part A is initially at rest a distance R away from the sun. What is the critical value of a
Sati [7]

Ans;

6.25m²/Kd

Explanation: see attached file

5 0
2 years ago
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