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Anarel [89]
3 years ago
13

From Doppler shifts of the spectral lines in the light coming from the east and west edges of the Sun, astronomers find that the

radial velocities of the two edges differ by about 4 km/s, meaning that the Sun’s rotation rate is 2 km/s. Find the approximate period of rotation of the Sun in days. The circumference of a sphere is given by 2πR, where R is the radius of the sphere
Physics
1 answer:
Gala2k [10]3 years ago
3 0

Answer:

T= 37 day

Explanation:

To solve this exercise we will use the definition of angular velocity as the angular distance, which for a full period is 2pi between time.

    w = T / t

The relationship between angular and linear velocity is

    v = w r

    w = v / r

We substitute everything in the first equation

    v / r = 2π / t

    t = 2π r / v

Let's reduce to the SI system

    V = 2 km / s (1000m / 1km) = 2 10³ m / s

    r= R = 6.96 10⁸ m

Let's calculate

    t = 2π 6.96 10⁸/2 10³

    t = 3.2 10⁶ s

    T = t = 3.2 10⁶ s ( 1h/3600s) (1 day/24 h)

    T= 37 day

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Answer:

\boxed{ \bold{ \huge{ \boxed {\sf{8.33 \: m \:/s \: }}}}}

Explanation:

Distance travelled = 200 metre

Time taken = 24 second

Velocity = ?

<u>Finding </u><u>the</u><u> </u><u>velocity</u><u> </u>

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\dashrightarrow{ \sf{velocity = 8.33 \: m/s}}

Hope I helped!

Best regards!

6 0
3 years ago
An electron enters the gap between the plates of a capacitor at the center of the gap traveling parallel to theplates at 2.0 x 1
Svetlanka [38]

Answer:

How far will the electron travel beforehitting a plate is 248.125mm

Explanation:

Applying Gauss' law:

Electric Field E = Charge density/epsilon nought

Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12

Therefore E = 1.0 x 10^-6/8.85× 10^-12

E= 1.13×10^5N/C

Force on electron F=qE

Where q=charge of electron=1.6×10^-19C

Therefore F=1.6×10^-19×1.13×10^5

F=1.808×10^-14N

Acceleration on electron a = Force/Mass

Where Mass of electron = 9.10938356 × 10^-31

Therefore a= 1.808×10^-14 /9.11 × 10-31

a= 1.985×10^16m/s^2

Time spent between plate = Distance/Speed

From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2

Therefore Time = 0.01/2×10^6

Time =5×10^-9s

How far the electron would travel S =ut+ at^2/2 where u=0

S= 1.985×10^16×(5×10^-9)^2/2

S=24.8125×10^-2m

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Answer:

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2. Xét một điểm nằm trên vành ngoài của lốp xe máy cách trục bánh xe môtô 25cm.
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Explanation:

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