Which component of a galaxy is often found between the stars and looks like a cloud or smoke?

A mass of 10.0 kg is in a gravitational field of 3.50 N/kg. What force acts on the mass?

mash [69]

Answer:

Force=35 N

Explanation:

Given data

mass m=10.0 kg

Gravitational field E=3.50 N/kg

To find

Force

Solution

From definition of gravitational field intensity.

$E_{gravitational-field }=\frac{Force}{mass}\\ E=F/m\\F=mE\\F=(10.0 kg)*(3.50 N/kg)\\F=35N$

6 0

4 years ago

Read 2 more answers

Lightning flashed, and after 7 seconds there was a roll of thunder. To the nearest meter, determine how far lightning flashed fr

Nadya [2.5K]

Let’s come back to the equation of SPEED OF AN ECHO
SPEED =2XDISTANCE/ TIME
So 337*7/2=1179.5 nearest meter =1800 m away
Hope you will get it right

3 0

3 years ago

Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars

zhuklara [117]

Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is $1.50 \times 10^{11}\;\rm m$ .

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

$T^2 \propto R^3$

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

$T^2 \propto R^3$

$T^2 = kR^3$

Substituting the values, we get the value of constant k for mars.

$687^2 = k\times (2.279 \times 10^{11})^3$

$k = 3.92 \times 10^{-29}$

The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

$365^3 = 3.92\times 10^{-29} R^3$

$R^3 = 3.39 \times 10^{33}$

$R= 1.50 \times 10^{11}\;\rm m$

Hence we can conclude that the distance of the earth from the sun is $1.50 \times 10^{11}\;\rm m$ .

To know more about Kepler's third law, follow the link given below.

brainly.com/question/7783290.

6 0

3 years ago

An 12 N force is applied to a 1 kg object. What is the magnitude of the objects acceleration?

Sauron [17]

Answer:

a=12 m/s²

Explanation:

Newton's second law of motion states that the acceleration of a body is directly proportional to the force applied and takes place in the direction of force.

This can be summarized as: F=ma, where m is the mass of the object on which force F acts. a is the acceleration due to the force applied.

12N= 1kg×a

a=12N/1kg

a=12m/s²

6 0

4 years ago

Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with the

scoundrel [369]

Answer:

F = 0.1575 N

Explanation:

When the third sphere touches the first sphere, the charge is distributed between both spheres, then now the first sphere has only half of his original charge.

In this moment then

Sphere one has a charge = Q/2

Sphere three has a charge = Q/2

Now when the third sphere touches the second sphere again the charge is distributed in a manner that both sphere has the same charge.

How the total charge is Q = Q/2 + Q = 3/2Q, when the spheres are separated each one has 3/4Q

Sphere two has a charge = 3/4Q

Sphere three has a charge = 3/4Q

The electrostatic force that acts on sphere 2 due to sphere 1 is:

F = $\frac{kQ_{1}Q_{2} }{r^{2} }$