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2 years ago

How do we know the sun rotates?

2 answers:

6 0

Hi pupil Here's your answer ::::

➡➡➡➡➡➡➡➡➡➡➡➡➡

The sun Doesnt rotates as we know all . But it rotates with the wjole our Milky Way Galaxy. This movement is so slow that we cant even recognise that we are shifted.

This we know because we all had studied about the solar system and space.

By this study we can say that Sun Rotates.

⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅

Hope this helps ......

oee [108]2 years ago

6 0

Answer: One side is blue-shifted and the other side is red-shifted.

Explanation: Correct Apex Answer 12/31/19

You might be interested in

Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon

aleksklad [387]

Answer:

230 N

Explanation:

At the lowest position , the velocity is maximum hence at this point, maximum support force T is given by the branch.

The swinging motion of the ape on a vertical circular path , will require

a centripetal force in upward direction . This is related to weight as follows

T - mg = m v² / R

R is radius of circular path . m is mass of the ape and velocity is 3.2 m/s

T = mg - mv² / R

T = 8.5 X 9.8 + 8.5 X 3.2² / .60 { R is length of hand of ape. }

T = 83.3 + 145.06

= 228.36

= 230 N ( approximately )

5 0

2 years ago

Two neutral metal spheres on wood stands are touching. A negatively charged rod is held directly above the top of the left spher

FrozenT [24]

Answer: Option (C) is the correct answer.

Explanation:

As we know that metals are able to conduct electricity so, when a negatively charges rod is kept closer to the left sphere then electrons will enter the sphere.

Since, like charges repel each other. Hence, some of the negative changes from the rod will repel the negative charges of left sphere.

As both left and right spheres are touching each other so, the electrons will move towards the right sphere. As a result, there will be too many electrons (negative charge) present on the right sphere and very less electrons present in the left sphere.

Thus, we can conclude that the statement right sphere is negatively charged, another is charged positively, is true.

7 0

2 years ago

Read 2 more answers

5 points

olga2289 [7]

Answer:

d. 5 ohms

Explanation:

For resistors in parallel, the equivalent resistance is found with:

1/Req = ∑(1/R)

1/R = 1/15 + 1/15 + 1/15

1/R = 3/15

R = 15/3

R = 5

8 0

3 years ago

A 129-kg horizontal platform is a uniform disk of radius 1.51 m and can rotate about the vertical axis through its center. A 67.

blondinia [14]

Answer:

I = 274.75 kg-m²

Explanation:

given,

mass of horizontal platform = 129 Kg

radius of the disk = 1.51 m

mass of the person(m_p) = 67.5 Kg

standing at distance(r_p) = 1.09 m

mass of dog(m_d) = 25.3 Kg

sitting near the person(r_d) = 1.37 m from center

moment of inertia = ?

Moment of inertia of disc = $\dfrac{1}{2}MR^2$

Moment of inertia of the system

$I = MOI\ of\ disk + m_p r_p^2 + m_d r_d^2$

$I = \dfrac{1}{2}MR^2 + m_p r_p^2 + m_d r_d^2$

$I = \dfrac{1}{2}\times 129 \times 1.51^2 + 67.5 \times 1.09^2 + 25.3\times 1.37^2$

I = 274.75 kg-m²

3 0

3 years ago

Question 1 (13 marks) A charge Q is located on the top-left corner of a square. Charges of 2Q, 3Q and 4Q are located on the othe

Colt1911 [192]

Answer:

Explanation:

First of all we shall calculate electric field near charge 2Q .

electric field due to charge Q = K x Q / (5 x 10⁻² )²

E₁ = KQ / 25 x 10⁻⁴ = KQ x 10⁴ / 25 . It is acting along positive x axis

E₁ = KQ x 10⁴ i / 25

Similarly electric field due to charge 3Q near 2Q

= 3KQ x 10⁴ i / 25 . It is acting along y-axis

E₂ = 3KQ x 10⁴ j / 25

Similarly electric field due to charge 4Q near 2Q

= 4KQ x 10⁴ j / (25 x 2 )

= 2 KQ x 10⁴ / 25 . It is acting acting along north east direction

unit vector in north east direction = ( i + j )/ √2

So E₃ can be represented by

E₃ = 2 KQ x 10⁴ ( i + j ) / 25 x √2

Total field = KQ x 10⁴ i / 25 + 3KQ x 10⁴ j / 25 + 2 KQ x 10⁴ ( i + j ) / ( 25 x √2 )

= KQ x 10⁴ [ i + 3 j + √2 i + √2 j ) / 25

= 400 KQ ( 2.414 i + 4.414 j ) N / C

Force on 2Q = Field x charge = 400 KQ ( 2.414 i + 4.414 j ) x 2Q N

= 800 KQ² ( 2.414 i + 4.414 j ) N

= 800 x 9 x 10⁹ x ( 2.5 x 10⁻⁶ )² x 2.414 x ( i + 2 j ) N

= 108.63 ( i + 2 j ) N .

Magnitude of this force

= 108.63 x √5

= 243 N approx .

4 0

3 years ago