3 years ago

What is the ratio of escape speed from earth to circular orbital speed? ignore air resistance.

Physics

2 answers:

klio [65]3 years ago

8 0

About 40 000 km/h
Here you go:

aleksklad [387]3 years ago

5 0

This would be my answer to the problem

You might be interested in

A particle of charge 3.53×10 −8 C experiences a force of magnitude 6.03×10 −6 N when it is placed in a particular point

Cloud [144]

<h2>Electric field at the location of the charge is 169.97 N/C</h2>

Explanation:

Electric field is the ratio of force and charge.

Force, F = 6 x 10⁻⁶ N

Charge, q = 3.53 x 10⁻⁸ C

We have

$E=\frac{F}{q}\\\\E=\frac{6\times 10^{-6}}{3.53\times 10^{-8}}\\\\E=169.97N/C$

Electric field at the location of the charge is 169.97 N/C

6 0

3 years ago

In an inelastic collision, which of the following statements is always true?

xz_007 [3.2K]

Well C is definitely one of the correct answers.

8 0

3 years ago

A cyclist reduces his speed from 6.5 m/s to 0.0 m/s with an acceleration of

kolbaska11 [484]

Answer:

a= -1.2 m/s^2

Vi= 6.5 m/s

Vf= 0 m/s

t= 0-6.5/-1.2= <u>5.45 Sec</u>

Explanation:

4 0

2 years ago

A student wants to determine the effect of mass on kinetic energy. She will drop two balls of the same size into a pool of water

Anastasy [175]

Answer:

She should drop two balls with different masses from the same height

Explanation:

4 0

2 years ago

Two charged objects are 1 meter apart. Calculate the magnitude of the electric force between them if the two charges are +1.0 μC

Solnce55 [7]

Answer:

0.00899 N

Explanation:

The magnitude of the electrostatic force between two charges is given by the equation:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the charges

r is the distance between the two charges

And the force is:

- Repulsive if the two charges have same sign

- Attractive if the two charges have opposite sign

In this problem we have:

$q_1=1.0\mu C = 1.0\cdot 10^{6}C$ (charge of object 1)

$q_2=1.0\mu C = 1.0\cdot 10^{6}C$ (charge of object 2)

r = 1 m (separation between the objects)

So, the electric force is

$F=(8.99\cdot 10^9)\frac{(1.0\cdot 10^{-6})^2}{1^2}=0.00899 N$

5 0

3 years ago