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Dmitry_Shevchenko [17]
3 years ago
12

A girl and a boy are riding on a merry-go-round that is turning at a constant rate. The girl is near the outer edge, and the boy

is closer to the center. For a given elapsed time interval, which rider has greater angular displacement?
A. Both the girl and the boy have the same nonzero angular displacement.
B. Both the girl and the boy have zero angular displacement.
C. The boy has greater angular displacement.
D. The girl has greater angular displacement.
Physics
1 answer:
masha68 [24]3 years ago
4 0

Answer:

A. Both the girl and the boy have the same nonzero angular displacement.

Explanation:

Since girl and boy both are standing on the same merry go round so here we can say that boy and girl both will revolve with same angular speed as the speed of the merry go round

so the angular displacement is given as

\theta = \omega t

so here since both has same angular speed that is the speed of the disc so they both will turn by same angle in the same interval of time

So we have correct answer as

A. Both the girl and the boy have the same nonzero angular displacement.

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A centrifuge used in DNA extraction spins at a maximum rate of 7000rpm producing a "g-force" on the sample that is 6000 times th
defon

Answer:

A) a = 73.304 rad/s²

B) Δθ = 3665.2 rad

Explanation:

A) From Newton's first equation of motion, we can say that;

a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.

Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s

Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s

We are given; t = 10 s

Thus;

a = 733.04/10

a = 73.304 rad/s²

B) From Newton's third equation of motion, we can say that;

ω² = ω_o² + 2aΔθ

Where Δθ is angular displacement

Making Δθ the subject;

Δθ = (ω² - ω_o²)/2a

At this point, ω = 0 rad/s while ω_o = 733.04 rad/s

Thus;

Δθ = (0² - 733.04²)/(2 × 73.304)

Δθ = -537347.6416/146.608

Δθ = - 3665.2 rad

We will take the absolute value.

Thus, Δθ = 3665.2 rad

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3 years ago
Convert centimeters to metre?
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Answer:1 cm=0.01

Explanation: Uu divide

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A train, traveling at a constant speed of 25.8 m/s, comes to an incline with a constant slope. While going up the incline, the t
zhannawk [14.2K]

The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

Answer:

Explanation:

So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.

So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then

v = 25.8 + (-1.66×8.3)

v =12.022 m/s.

So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

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A typical jet airliner has a cruise airspeed of 900 km/h , which is its speed relative to the air through which it is flying. If
Luba_88 [7]

Explanation:

It is given that,

Speed of the jet airplane with respect to air, v_{PA} = 900\ km/h          

If the wind at the airliner’s cruise altitude is blowing at 100 km/h from west to east, v_{AG}= 100\ km/h

(A) Let v_{PG} is the speed of the airliner relative to the ground if the airplane is flying from west to east,

v_{PG}=900-100=800\ km/h

(B) Let v'_{PG} is the speed of the airliner relative to the ground if the airplane is flying from east to west,

v'_{PG}=900+100=1000\ km/h

Hence, this is the required solution.                                            

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The answer is left and right. 
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