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3 years ago

Which statement relates most directly to the second law?

Physics

2 answers:

alukav5142 [94]3 years ago

7 0

Statement three i do believe

jok3333 [9.3K]3 years ago

6 0

Answer: Option (2) is the correct answer.

Explanation:

According to second law of thermodynamics, all the energy of a system cannot be converted into useful work because some amount of energy will be lost.

For example, a machine will never convert its energy into 100% work as some of the energy of machine will be lost into the atmosphere.

Therefore, we can conclude that the statement some thermal energy is lost and some is used to do work, relates most directly to the second law.

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Can we use position as another word for phase?

Yuri [45]

Answer:

yes

Explanation:

6 0

2 years ago

A constant friction force of 25 N acts on a 65-kg skier for 15 s on level snow. What is the skier’s change in velocity

nikdorinn [45]

Answer:

$\Delta v=5.77m/s$

Explanation:

Newton's 2nd Law relates the net force F on an object of mass m with the acceleration a it experiments by F=ma. In our case the net force is the friction force, since it's the only one the skier is experimenting horizontally and the vertical ones cancel out since he's not moving in that direction. Our acceleration then will be:

$a=\frac{F}{m}$

Also, acceleration is defined by the change of velocity $\Delta v$ in a given time t, so we have:

$a=\frac{\Delta v}{t}$

Since we want the change in velocity, mixing both equations we conclude that:

$\Delta v=at=\frac{Ft}{m}$

Which for our values means:

$\Delta v=\frac{Ft}{m}=\frac{(25N)(15s)}{(65Kg)}=5.77m/s$

4 0

3 years ago

An insect crawls 4 m east along a hiking path in

alexandr1967 [171]

The insect's velocity is 0.2 meter/second along east.

<h3>What is velocity?</h3>

The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction. SI unit of velocity is meter/second.

Given parameters:

An insect crawls 4 m east along a hiking path.

Time taken by it: t = 20 seconds.

We have to fine: the insect's velocity, v =?

So, the magnitude of velocity: v= 4/20 meter/second = 0.2 meter/second.

And the direction velocity is along east.

Hence, the insect's velocity is 0.2 meter/second along east.

Learn more about velocity here:

brainly.com/question/18084516

#SPJ1

8 0

1 year ago

QUICKLY I AM IN THE LESSON NOWWW

vladimir2022 [97]

Answer:

I don't really know much about the whole stars thing, but I've written a lot of paragraphs in science

Explanation:

Remember:

A paragraph should be in CER formation: Claim (hypothesis/discovery), Evidence, and Reasoning
Your reasoning should never start with I believe or I think, your reasoning and evidence are 100% fact based
The actual hypothesis should only be 1-2 sentences and be straight to the point

Include:

for your hypothesis, keywords relating to your subject, I'm assuming for your topic you should include things like marking, telescope, stars, etc.
why you believe your hypothesis is true
so use prior knowledge to justify your hypothesis

7 0

2 years ago

The spring of modulus k = 200 n /m is compressed a distance of 300 mm and suddenly released with the system at rest. determine t

DiKsa [7]

I attached the missing picture.
Let's analyze the situation as spring goes from stretched to unstretched state.
When you stretch the string you have to use force against ( you are doing work) this energy is then stored in the spring in the form of potential energy. When we release the spring the energy is being used to push the two carts. When the spring reaches its unstretched length its whole initial potential energy has been used on the carts, and this is the moment when two carts have maximum velocity.
The potential energy of compressed ( stretched) spring is:
$E_p=\frac{1}{2}kx^2$
The kinetic energy of two carts is:
$E_{k1}+E_{k2}=m_1\frac{v_1^2}{2}+m_2\frac{v_2^2}{2}$
So we have:
$E_p=E_{k1}+E_{k2}\\ \frac{1}{2}kx^2=m_1\frac{v_1^2}{2}+m_2\frac{v_2^2}{2}$
Momentum also has to be conserved:
$m_1v_1-m_2v_2=0\\ m_1v_1=m_2v_2\\ v_1=\frac{m_2}{m_1}v_2$
Momentum before the release of the spring is zero so it has to stay zero. We plug this back into the expresion we got from law of conservation of energy and we get:
$v_2^2=\frac{m_1^2}{m_2^2-m_1^2}kx^2=4.05\\
v_2=\sqrt{4.05}=2.012\frac{m}{s}$
Now we go back to the momentum equation:
$v_1=\frac{m_2}{m_1}v_2\\
v_1=4.69\frac{m}{s}$

8 0

3 years ago