Ask question

Ask question

All categories

Misha Larkins [42]

3 years ago

12

what is the displacement of the earth after one orbit about the the sun? what is the average velocity of the earth after one orb

it in m/s

1 answer:

valina [46]3 years ago

6 0

5.7 seconds idk idk

You might be interested in

Your spaceship lands on an unknown planet. to determine the characteristics of this planet, you drop a 1.50 kg wrench from 5.50

Vesnalui [34]

1. calculate the value of acceleration that objects gains in that period of time
•calculating acceleration
5.50 = 1/2at^2
5.50*2/t^2 = a
11.00/0.657 = a
16.74=a
now you got the acceleration
2. you have laws of gravitation for that

g = Gm/r^2
where g is the acceleration value
16.74 = 6.754*10^-11 × m/ 6.28*10^4
105.14*10^4 /6.754*10-11 = m
15.567*10^15 = m
that would be the mass of the planet ...

5 0

3 years ago

Read 2 more answers

1. Which wave phenomenon is illustrated by this image?

Arisa [49]

Answer:

Explanation: O B. Diffraction

7 0

1 year ago

Read 2 more answers

!!PLEASE ANSWER ASAP!!

galina1969 [7]

A. elastic motion because that's the answer

8 0

2 years ago

Read 2 more answers

A quarterback claims that he can throw the football a horizontal distance of 167 m. Furthermore, he claims that he can do this b

pychu [463]

Answer:u=42.29 m/s

Explanation:

Given

Horizontal distance=167 m

launch angle $=33.1^{\circ}$

Let u be the initial speed of ball

Range $=\frac{u^2\sin 2\theta }{g}$

$167=\frac{u^2\sin (66.2)}{9.8}$

$u^2=1788.71$

$u=\sqrt{1788.71}$

$u=42.29 m/s$

7 0

2 years ago

3. What is the gravitational force between a 70 kg physics student and her 1 kg textbook, at a distance of 1 meter? (This number

Rina8888 [55]

ANSWER

$\begin{equation*} 4.67*10^{-9}\text{ }N \end{equation*}$

EXPLANATION

Parameters given:

Mass of the student, M = 70 kg

Mass of the textbook, m = 1 kg

Distance, r = 1 m

To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:

$F=\frac{GMm}{r^2}$

where G = gravitational constant

Therefore, the gravitational force acting between the student and the textbook is:

$\begin{gathered} F=\frac{6.67430*10^{-11}*70*1}{1^2} \\ \\ F=4.67*10^{-9}\text{ }N \end{gathered}$

That is the answer.

6 0

1 year ago