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zmey [24]
3 years ago
6

Why is your velocity continuously changing as you ride on a carousel?

Physics
2 answers:
Finger [1]3 years ago
5 0

Since carousels go in circles, the direction of travel is constantly changing and velocity is defined as constant change. This is the reason why velocity is continuously changing when riding a carousel for velocity is involve in change of direction.

Moreover, a change in velocity is an acceleration and velocity has two properties that comprises direction and magnitude. Whereas the direction of velocity is the direction of travel and speed is the magnitude of velocity.

nydimaria [60]3 years ago
5 0
<span>To make this more obvious (and give it some relevance beyond being just a HW question), imagine that the ride is going super, super, duper fast. You'd better hold on tight or else you'll fly off. There's obviously a force acting on you, then, to keep you on that ride. Therefore, you are accelerating, which means your velocity is changing.</span>
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What would happen to the wavelength if the frequency was doubled?
ira [324]

Answer:

λ = hv

If frequency is doubled :

λ = h × 2v

λ = 2hv

Thus wavelength is doubled

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3 years ago
A person drops a brick from the top of a building. The height of the building is 400 m and the mass of the brick is 2.00 kg. Wh
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3 years ago
According to the diagram of the electromagnetic spectrum shown, what would best represent the
iren2701 [21]

The  best represent the size of visible light will be C. Protozoa

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the best represent the size of visible light is Protozoa, According to the diagram of the electromagnetic spectrum shown,

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6 0
8 months ago
A wheel with moment of inertia 25 kg. m2 and angular velocity 10 rad/s begins to speed up, with angular acceleration 15 rad/sec2
Pani-rosa [81]

Answer:

(A) Angular speed 40 rad/sec

Rotation = 50 rad

(b) 37812.5 J

Explanation:

We have given moment of inertia of the wheel I=25kgm^2

Initial angular velocity of the wheel \omega _0=10rad/sec

Angular acceleration \alpha =15rad/sec^2

(a) We know that \omega =\omega _0+\alpha t

We have given t = 2 sec

So \omega =10+15\times  2=40rad/sec

Now \Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad

(b) After 3 sec \omega =10+15\times 3=55rad/sec

We know that kinetic energy is given by Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J

7 0
3 years ago
A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The ang
olga55 [171]

Answer:

9.22 s

Explanation:

One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad

Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:

For the child: s_c = \omega_ct = 0.233t

For the horse: s_h = s_0 + a_ht^2/2 = 1.57 + 0.0136t^2/2 = 1.57 + 0.0068t^2

For the child to catch up with the horse, they must cover the same angular distance within the same time t:

s_c = s_h

0.233t = 1.57 + 0.0068t^2

0.0068t^2 - 0.233t + 1.57 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{0.233\pm \sqrt{(-0.233)^2 - 4*(0.0068)*(1.57)}}{2*(0.0068)}

t= \frac{0.233\pm0.11}{0.0136}

t = 25.05 or t = 9.22

Since we are looking for the shortest time we will pick t = 9.22 s

6 0
3 years ago
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