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zloy xaker [14]
3 years ago
13

What is bricks matter rigidity ​

Physics
1 answer:
expeople1 [14]3 years ago
3 0

Answer:

In a tabular column represent the following characteristics of matter-rigidity, compressibility, shape, kinetic energy and density of following substances- brick, honey, air, block of wood, water

Explanation:

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The earth exerts a force of 1900 N on an or-
Aleks [24]

The satellite exerts a force of 1900 N on the Earth

Explanation:

To answer this question, let's remind Newton's third law:

"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"

In this problem, we can identify:

The Earth as object A

The satellite as object B

This means that if we apply the law, the force that the Earth exerts on the satellite must be equal (in magnitude) and opposite (in direction) to the force the satellite exerts on the Earth.

Since the Earth exerts a force of 1900 N on the satellite, this means that the satellite also exerts a force of 1900 N on the Earth.

Learn more about Newton's third law:

brainly.com/question/11411375

#LearnwithBrainly

5 0
3 years ago
You're in your room blasting music with door shut, your mom opens your door. Now music is heard through out your home. this is e
Sliva [168]

Answer:

A

Explanation:

Diffraction, as the waves spread out (specifically spread to the whole house) by passing the door.

5 0
3 years ago
A body of mass 2 kg at O has an initial velocity of 3m/s along OE and it is subjected to a force of 4N perpendicular to OE the d
timofeeve [1]
Perpendicular acceleration:
F = ma
a = 4 / 2 = 2 m/s²

Perpendicular distance:

s = ut + 1/2 at²
s = 0 x 4 + 1/2 x 2 x 4²
s = 16 m

Horizontal distance:
s = ut
= 3 x 4
= 12 m

Total distance = √(12² + 16²)
= 20 m.
3 0
4 years ago
Evaluate x and y in the equation: E=Cm^xV^y , where E is kinetic energy , m is mass , V is velocity and C is a dimension less co
Korvikt [17]

Answer: Do I look like Einstein

8 0
3 years ago
If a galaxy has an apparent velocity of 2300 km/s, what is its distance if the Hubble constant is assumed to be 70 km/s/Mpc
prohojiy [21]

The distance of the galaxy is 32.86 Mpc.

Using the hubble law, v = H₀D where v = apparent velocity of galaxy = 2300 km/s, H = hubble constant = 70 km/s/Mpc and D = distance of galaxy.

Since we require the distance of the galaxy, we make D subject of the formula in the equation. So, we have

D = v/H₀

Substituting the values of the variables into the equation, we have

D = 2300 km/s ÷ 70 km/s/Mpc

D = 32.86 Mpc

So, the distance of the galaxy is 32.86 Mpc

Learn more about hubble law here:

brainly.com/question/18484687

4 0
2 years ago
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