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Irina18 [472]
3 years ago
12

Which phrases describe all the outer planets’ motion? Select two options.

Physics
2 answers:
AysviL [449]3 years ago
8 0

Answer:

Fast rotation

Slow revolution

Explanation:

Solar system has 8 planets. 4 inner rocky planets - Mercury, Venus, Earth and Mars and 4 outer gaseous planets - Jupiter, Saturn, Uranus and Neptune.  The outer planets have few common features.

They are gaseous. There  period of revolution is larger than the inner planets which means that they have slow revolution about the Sun. One day on the outer planets is smaller than the inner planets which means they have fast rotation.

For example, Jupiter has revolves around sun in 11.86 Earth years and rotates about axis in 9.8 Earth hours. Uranus revolves around sun in 84 Earth years and rotates on its axis 17.9 Earth hours.

Ainat [17]3 years ago
4 0

Answer:

Fast rotation

Slow revolution

Explanation:

did the test

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if the speed of a ball increased from 1m/s to 4 m/s, by how much would kinetic energy increase? .....help me please.
ivann1987 [24]
Can you write the formula for kinetic energy ?
Here, let me help you:

       Kinetic energy = (1/2) (mass) (speed)²  .

Look at the (speed) in the formula.  It's squared.

So if the speed gets multiplied by (something),
the kinetic energy gets multiplied by  (something)² .

If the speed starts out at 1 m/s, but gets multiplied by 4,
then the kinetic energy gets multiplied by  (4)²  =  16 .
7 0
4 years ago
A current of 16.0 mA is maintained in a single circular loop of 1.90 m circumference. A magnetic field of 0.790 T is directed pa
pav-90 [236]

Answer

given,

current (I) = 16 mA

circumference of the circular loop (S)= 1.90 m

Magnetic field (B)= 0.790 T

S = 2 π r

1.9 = 2 π r

r = 0.3024 m

a) magnetic moment of loop

    M= I A

    M=16 \times 10^{-3} \times \pi \times r^2

   M=16 \times 10^{-3} \times \pi \times 0.3024^2

   M=4.59 \times 10^{-3}\ A m^2

b)  torque exerted in the loop

 \tau = M\ B

 \tau = 4.59 \times 10^{-3}\times 0.79

 \tau = 3.63 \times 10^{-3} N.m

8 0
3 years ago
A circular hoop of diameter d hangs on a nail. what is the period of its oscillations at small amplitude?
Misha Larkins [42]

m = mass of the circular hoop

r = radius of the hoop

I = moment of inertia of the hoop

moment of inertia of the hoop about the center of hoop is given as

I = m r²

k = distance of the point of suspension from center of mass = r

using parallel axis theorem

I' = moment of inertia of hoop about the point of suspension

I' = I + m k²

I' = m r² + m k²

I' = m r² + m r²

I' = 2 m r²

Time period of oscillation for the hoop is given as

T = 2π sqrt(I'/mgk)

T = 2π sqrt(2 m r²/(mgr))

T = 2π sqrt(2 r/g)

since 2r = diameter = d

T = 2π sqrt(d/g)



4 0
4 years ago
A car capable of a constant acceleration of
Sav [38]

Answer:

sorry we don't understand what about of your question is

Explanation:

sorry we don't have a great weekend

6 0
4 years ago
Cho các máy cắt sử dụng trong công nghiệp có ký hiệu trên nhãn thiết bị: C350; B500. Hãy tính dòng điện bảo vệ ngắn mạch và dòng
solniwko [45]

Answer:

ask in the English then I can help you

Explanation:

please mark me as brain list

3 0
3 years ago
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