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3 years ago

Which of Kepler's laws describes the speed at which planets travel at different points in their orbit around the Sun?

Physics

2 answers:

marishachu [46]3 years ago

5 0

Answer: D) The law of equal area and equal time

Explanation:

vagabundo [1.1K]3 years ago

4 0

D. This says the path sweeps out equal segments of the elliptical path in equal times (picture a wedge of a pie shaped like an oval)

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What is the potential energy of a 2kg plant that is on a windowsill 1.3 m high?

klio [65]

Answer:

25.48

Explanation:

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3 years ago

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Ulleksa [173]

Explanation:

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2 years ago

Read 2 more answers

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges

Nostrana [21]

Answer:

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

$E_1 = E_2$

Let the position where net field is zero will lie at distance "r" from q1

$\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}$

now we will have

$\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}$

now square root both sides

$\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}$

now we have

$\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1$

so we have

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

8 0

3 years ago

you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k

alex41 [277]

Answer:

You should choose airspeed 158.11 km/h at 18.4° west of south

Explanation:

The distance to the air port is 450 km due to south

You should to reach the airport in 3 hours

→ Velocity = distance ÷ time

→ Distance = 450 km , time = 3 hours

→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south

A wind is blowing from west 50 km/h

We need to know what heading and airspeed you should choose to

reach your destination

At first we must find the resultant velocity of your plane and the wind

The south and west are perpendicular, then the resultant velocity is

→ $v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}$

→ $v_{p}=150$ km/h , $v_{w}=50$ km/h

→ $v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11$ km/h

To cancel the velocity of the wind, the pilot should maintain the velocity

of the plane at 158.11 km/h

The direction of the velocity is the angle between the resultant velocity

and the vertical (south)

→ The direction of the velocity is $tan^{-1}\frac{50}{150}=18.4$ °

The direction of the velocity is 18.4° west of south

<em>You should choose airspeed 158.11 km/h at 18.4° west of south</em>

8 0

3 years ago

Kon'nichiwa~<br>please help me with this question!!

andrezito [222]

Let's check the relationship

$\\ \rm\hookrightarrow g=\dfrac{GM}{r^2}$

$\\ \rm\hookrightarrow g\propto G$

Raindrops will fall faster . .
Also walking on ground would become more difficult as g increases.

Option C is wrong by now .Let's check D once

$\\ \rm\hookrightarrow T\propto \dfrac{1}{\sqrt{g}}$

So time period of simple pendulum would decrease.

4 0

2 years ago