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3 years ago

What is the centripetal force that holds planets in orbit?

Physics

1 answer:

Fudgin [204]3 years ago

8 0

<h2>Answer: Gravity force</h2>

If we approximate the orbit of the planets around the Sun to circular orbits with a uniform circular motion, where the velocity $\vec{V}$ is a vector, whose direction is perpendicular to the radius $r$ of the trajectory; the acceleration $\vec{a}$ is directed towards the center of the circumference (that's why it's called centripetal acceleration).

Now, according to Newton's 2nd law, the force $\vec{F}$ is directly proportional and in the same direction as the acceleration:

$\vec{F}=m.\vec{a}$

Therefore the net force resulting from the movement of a planet orbiting the Sun points towards the center of the circle, this is called Centripetal Force which is a central force that in this case is equal to the gravity force.

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Three equal point charges, each with charge 1.45 μCμC , are placed at the vertices of an equilateral triangle whose sides are of

LUCKY_DIMON [66]

Answer:

U = 80.91 J

Explanation:

In order to calculate the electric potential energy between the three charges you use the following formula:

$U=k\frac{q_1q_2}{r_{1,2}}$ (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q1: q2 charge

r1,2: distance between charges 1 and 2.

For the three charges you have:

$U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}$ (2)

You use the fact that q1=q2=q3=q and that the distance between charges are equal. Then, in the equation (2) you have:

q = 1.45μC = 1.45*10^-6C

r = 0.700mm = 0.700*10^-3m

$U_T=3k\frac{q^2}{r}=3(8.98*10^9Nm^2/C^2)\frac{(1.45*10^{-6}C)}{0.700*10^{-3}m}\\\\U_T=80.91J$

The electric potential energy between the three charges is 80.91 J

7 0

3 years ago

A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. If the plane is traveling hori

11111nata11111 [884]

Answer:

481 m

Explanation:

To fall 235 m, the time required is

t = √(2H/g)

t= √(2 $\times$ 235/9.8)

t=6.92 seconds.

The supplies will travel forward

6.92 $\times$ 69.4 ≈ 481 m

Therefore, the goods must be dropped 481 m in advance of the recipients.

5 0

2 years ago

How long must a 0.70-mm-diameter aluminum wire be to have a 0.40 a current when connected to the terminals of a 1.5 v flashlight

Natalka [10]

By using Ohm's law, we can find what should be the resistance of the wire, R:
$R= \frac{V}{I}= \frac{1.5 V}{0.40 A} =3.75 \Omega$

Then, let's find the cross-sectional area of the wire. Its radius is half the diameter,
$r=35 mm=0.35 \cdot 10^{-3} m$
So the area is
$A=\pi r^2 = \pi (0.35 \cdot 10^{-3} m)^2=3.85 \cdot 10^{-7} m^2$

And by using the resistivity of the Aluminum, $\rho=2.65 \cdot 10^{-8} \Omega m$ , we can use the relationship between resistance R and resistivity:
$R= \frac{\rho L}{A}$
to find L, the length of the wire:
$L= \frac{RA}{\rho}= \frac{(3.75 \Omega)(3,85 \cdot 10^{-7} m^2)}{2.65 \cdot 10^{-8} \Omega m}=54.48 m$

4 0

3 years ago

Choose the law each sentence describes. This law relates a planet's orbital period and its average distance to the Sun. The orbi

hram777 [196]

These are the Kepler's laws of planetary motion.

This law relates a planet's orbital period and its average distance to the Sun. - Third law of Kepler.

The orbits of planets are ellipses with the Sun at one focus. - First law of Kepler.

The speed of a planet varies, such that a planet sweeps out an equal area in equal time frames. - Second law of Kepler.

7 0

3 years ago

Read 2 more answers

I do not know where to start.

irakobra [83]

Underline the words then eliminate the ones that arent part of the problem!

4 0

3 years ago