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2 years ago

14

Accorrding to Kepler's third law, a planet whose distance from the Sun is 2 A.U. would have an orbital period of how many Earth-

years

1 answer:

Lilit [14]2 years ago

4 0

Answer:

hi

Explanation:

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A child is riding in a wagon. What reference frame might have been used if an observer said the child was not moving.

GarryVolchara [31]

Answer:

the wagon should be used as frame of reference if an observer said the child was not moving.

Explanation:

The state of motion of a body depends upon the frame of reference. It is the set of co-ordinates according to which the motion is analyzed. If a child is riding in a wagon, then he will be considered in motion to a person standing outside the wagon. Hence, if we take a frame of reference outside the wagon then the child must be in motion with respect to the observer. On the other hand if the observer is inside the wagon, then the child must be in rest with respect to the observer. Hence, if we take the wagon to be the frame of reference, then the child will be at rest with respect to the observer.

<u>Therefore, the wagon should be used as frame of reference if an observer said the child was not moving.</u>

3 0

2 years ago

Ahab has been sent to the shops by his mother. The shops are 1200 m from his

Iteru [2.4K]

Answer:

it will take him 200secs to run to the shops

(btw 200 seconds is 3minutes and 20secs)

Explanation:

distance = speed x time

time = distance / speed

1200 ÷ 6 = 200

time = 200seconds

hope this helps

brainliest please?

x

8 0

2 years ago

which pole of a compass needle points north ? please give simple answer with little explaining if any :)

Mumz [18]

I think the North Pole
If I understand this correctly

6 0

2 years ago

During World War II, mass spectrometers were used to separate the radioactive uranium isotope U-235 from its far more common iso

mariarad [96]

Answer:

21.55 m

Explanation:

6 0

2 years ago

A positive point charge Q is located at x=a and a negative point charge −Q is at x=−a. A positive charge q can be placed anywher

Scilla [17]

Answer:

$\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x$

Explanation:

The Coulomb's Law gives the force by the charges:

$\vec{F} = K\frac{q_1q_2}{r^2}\^r$

Let us denote the positon of the charge q on the y-axis as 'y'.

The force between 'Q' and'q' is

$F_1 = K\frac{Qq}{x^2 + y^2}\\F_1_x = F_1\cos(\theta)$

where Θ is the angle between $F_1$ and x-axis.

$F_1_x = K\frac{Qq}{x^2 + y^2}(\frac{x}{\sqrt{x^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}$

whereas

$F_2_x = K\frac{-Qq}{a^2 + y^2}(-\frac{a}{\sqrt{a^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}$

Finally, the x-component of the net force is

$\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x$

8 0

3 years ago