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Umnica [9.8K]
3 years ago
12

you and a friend sit still on a spinning merry-go-round. The merry-go-round spins 5 times every second, and to a stationary obse

rver, you have a speed of 9m/s. From your frame of reference, what is the speed of your friend?
Physics
2 answers:
iris [78.8K]3 years ago
6 0

Answer:

the other person is right, the answer is zero.

Explanation:

dedylja [7]3 years ago
4 0

Answer:

  • <u> From your frame of reference, the speed of your friend is zero.</u>

Explanation:

The<em> speed</em> from a <em>frame of reference</em> is the magnitude of the change of the position per unit of time in that frame of reference.

When it is said that, to a <em>stationary observer</em> you have a <em>speed of 9m/s</em>, it means that the observer "measures"  tha your position, in a moment, changes 9 m per second.

Since both you and your friend are sit still on the same spinning merry-go-round, neither of you will change positions with respect to the other, meaning that you will perceive that your friend will not move, from your frame of reference. That is, if you are isolated from the rest of the universe, and the merry-go-round continues spining uniformly, you cannot detect any movement of your friend, which means that from your frame of reference, the speed of your friend is zero.

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Spinning situations???
Papessa [141]
Just like mass, energy, linear momentum, and electric charge, angular momentum is also conserved.

The wheel has angular momentum.  I don't remember whether it's
up or down (right-hand or left-hand rule), but it's consistent with
counterclockwise rotation as viewed from above.

When you grab the wheel and stop it from spinning (relative to you),
that angular momentum has to go somewhere.

As I see it, the angular momentum transfers through you as a temporary
axis of rotation, and eventually to the merry-go-round. Finally, all the mass
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axis, counterclockwise as viewed from above, and with the magnitude
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7 0
3 years ago
Read 2 more answers
What would be the speed of an object just before hitting the ground if dropped 91.5 meters
Aleks04 [339]

Answer:

about 42.35 m/s

Explanation:

Use the equation for accelerated motion (g), and with zero initial velocity that doesn't include time:

v_f^2=v_i^2+2\,a\,\Delta x

which for our case would reduce to:

v_f^2=v_i^2+2\,a\,\Delta x\\v_f^2=0+2\,9.8\,(91.5)\\v_f^2= 1793.4\\v_f=\sqrt{1793.4} \\v_f \approx 42.35

then the velocity just before hitting would be about 42.35 m/s

5 0
2 years ago
A sound wave has a frequency of 295 Hz and travels the length of a football field, 91.4 m in 0.506 s. What is the period of the
dem82 [27]

Answer:1/295 seconds

Explanation:

Period=1/frequency

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7 0
3 years ago
Two ships leave a harbor at the same time, traveling on courses that have an angle of 110∘ between them. If the first ship trave
Allushta [10]

Answer:

49.07 miles

Explanation:

Angle between two ships = 110° = θ

First ship speed = 22 mph

Second ship speed = 34 mph

Distance covered by first ship after 1.2 hours = 22×1.2 = 26.4 miles = b

Distance covered by second ship after 1.2 hours = 34×1.2 = 40.8 miles = c

Here the angle between the two sides of a triangle is 110° so from the law of cosines we get

a² = b²+c²-2bc cosθ

⇒a² = 26.4²+40.8²-2×26.4×40.8 cos110

⇒a² = 2408.4

⇒a = 49.07 miles

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saveliy_v [14]
The answers are B, D, E!
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2 years ago
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