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Natasha2012 [34]
3 years ago
11

What term is used to describe a system that loses energy to objects that are not part of the system?

Physics
2 answers:
steposvetlana [31]3 years ago
5 0
<span>What term is used to describe a system that loses energy to objects that are not part of the system?

-Open. </span>
Mandarinka [93]3 years ago
3 0

Explanation:

A closed system is a system in which there is no exchange of matter and energy between the system and surrounding.

For example, thermos is a closed system as there is no exchange of matter and energy from thermos to the surrounding or vice versa.

Whereas when there is exchange of energy and matter between the system and surrounding then it is known as open system.

For example, a cup of coffee is an open system as energy (heat) from the coffee goes to the surrounding.

Thus, we can conclude that a system that loses energy to objects that are not part of the system is called an open system.

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Relative to the distance of an object in front of a plane mirror, how far behind the mirror is the image?
koban [17]
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Building helps children develope their _____ skills
Musya8 [376]

Answer:

I'll say number sense and operation

Explanation:

this is because it will show how creative and Intelligent that child is,and it will help increase thier brain functions

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3 years ago
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At the focal point of a converging lens, the object:
rewona [7]

Answer:

D) Image cannot be seen

Explanation:

As we know by lens formula

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

now we know that

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d_o = f

so we will have

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8 0
4 years ago
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Two soccer players, Mia and Alice, are running as Alice passes the ball to Mia. Mia is running due north with a speed of 5.30 m/
Tju [1.3M]

Answer:

a) v_{p}  = 2.83 m / s ,  b)  50.5º north east

Explanation:

This is a vector problem.

         v_{bg} = v_{bm} + v_{mg}

The speed of the ball with respect to the ground is the speed of the ball with respect to Mia plus the speed of Mia with respect to the ground

To make the sum we decompose the speed of the ball in its components

The angle of 30 east of the south, measured from the positive side of the x axis is

             θ = 30 + 270 = 300

            v_{bx} = v_{b} cos 300

             v_{by} = v_{b} sin. 300

             v_{bx} = 3.60 cos 300 = 1.8 m / s

             v_{by} = 3.60 sin 300 = -3,118 m / s

Let's add speeds on each axis

X axis

            vₓ = v_{bx}

             vₓ = 1.8 m / s

Y Axis  

             v_{y} = v1 - vpy

             v_{y} = 5.30 - 3.118

             v_{y} = 2.182 m / s

The magnitude of the velocity can be found using the Pythagorean theorem

              v_{p} = √ (vₓ² + v_{y}²)

               v_{p} = √ (1.8² + 2.182²)

               v_{p} = 2,829 m / s

               v_{p}  = 2.83 m / s

b) for direction use trigonometry

              tan θ = v_{y} / vₓ

              θ = tan ⁺¹ v_{y} / vₓ

              θ = tan⁻¹ 2.182 / 1.8

         Tea = 50.48º

This address is 50.5º north east

8 0
3 years ago
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