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liberstina [14]
3 years ago
11

I need help I'm stuck on a question it's a cross word "the measure of energy of motion of particles of matter

Physics
2 answers:
Masteriza [31]3 years ago
7 0
<span><span>Temperaturea measure of the average energy of random motion of particles of matter</span><span>Thermal Energythe total energy of all the particles of matter</span><span>Exothermic changea change in which energy is given off</span><span>Kinetic energy<span>the energy of matter in motion</span></span></span>
natima [27]3 years ago
3 0
When you ask for help with a crossword clue, don't forget to tell us how many letters it has. For this one, see if "temperature" fits the space.
You might be interested in
A force does 30000 J of work along a distance of 9.5m. Find the applied force.
Elina [12.6K]

Answer: F=3158N

Explanation:

Work is the product of force applied and the distance the object moves along the force applied. Work is measured in joules and the equation is as follows

W = F x d

So that F = W/d

F = 30000 J / 9.5m

F = ~3158 N

8 0
3 years ago
The masses of the Moon and Earth are 7.35 x 1022 kg and 5.97 x 1024 kg, respectively. The strength of the gravitational force be
bixtya [17]

Answer:

Distance between centre of Earth and centre of Moon is 3.85 x 10⁸ m

Explanation:

The attractive force experienced by two mass objects is known as Gravitational force.

The gravitational force is determine by the relation:

F=\frac{Gm_{1} m_{2} }{d^{2} }      ....(1)

According to the problem,

Mass of Moon, m₁ = 7.35 x 10²² kg

Mass of Earth, m₂ = 5.97 x 10²⁴ kg

Gravitational force experienced by them, F = 1.98 x 10²⁰ N

Universal gravitational constant, G = 6.67 x 10⁻¹¹ Nm²kg⁻²

Substitute these values in equation (1).

1.98\times10^{20} =\frac{6.67\times10^{-11}\times7.35\times10^{22}\times5.97\times10^{24} }{d^{2} }

d^{2}=\frac{2.93\times10^{37}}{1.98\times10^{20}}

d=\sqrt{1.48\times10^{17}}

d = 3.85 x 10⁸ m

3 0
3 years ago
I need help on this question!
Mashutka [201]

Explanation:

click the link you can get answer of the question

3 0
3 years ago
please solve this for me ,A garden roller is pulled with a force of 200N acting at an angle of 50 degree with the ground level.f
Bingel [31]

Answer:

The force pulling the roller along the ground is 128.55 N

Explanation:

A force of 200 N acting at an angle of 50° with the ground level

This force is pulled a garden roller

We need to find the force pulling the roller along the ground

The force that pulling the roller along the ground is the horizontal

component of the force acting

→ The force acting is 200 N at direction 50° with ground (horizontal)

→ The horizontal component = F cosФ

→ F = 200 N , Ф = 50

→ The horizontal component = 200 cos(50) = 128.55 N

128.55 N is the horizontal component of the force that pulling the

roller along the ground

<em>The force pulling the roller along the ground is 128.55 N</em>

8 0
3 years ago
HELP PLEASE!! An illustration of a pendulum at 3 positions of a pendulum. The equilibrium position is labeled B. The other two p
Sedbober [7]

Answer:

Both A and C

Explanation:

I just got it correct on Edg

7 0
2 years ago
Read 2 more answers
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