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MaRussiya [10]
3 years ago
11

When is thermal equilibrium achieved between two objects?

Physics
2 answers:
BARSIC [14]3 years ago
6 0

Answer:

C) When both objects have the same temperature

Explanation:

took the quiz

S_A_V [24]3 years ago
3 0
I believe the correct answer from the choices listed above is option C. Thermal equilibrium is achieved between two objects when both objects have the same temperature. <span>This is the condition under which two substances in physical contact with each other exchange no heat energy. Hope this answers the question.</span>
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On a spacecraft, two engines are turned on for 684 s at a moment when the velocity of the craft has x and y components of v0x 43
Natasha2012 [34]

As we know by the kinematics

\delta x = v_x * t + \frac{1}{2} at^2

4.11106 = 4370* 684 + \frac{1}{2}a_x*684^2

4.11106 = 2989080 + 233928*a_x

a_x = -12.78 m/s^2

Similarly for Y direction

\delta y = v_y * t + \frac{1}{2} at^2

6.07106 = 6280* 684 + \frac{1}{2}a_y*684^2

6.07106 = 4295520 + 233928*a_y

a_y = -18.36 m/s^2

8 0
3 years ago
Read 2 more answers
If one knows only the constant resultant force acting on an object and the time during which this force acts
Alenkinab [10]

Explanation:

  • We know that Newton's second law of motion gives the measure of external force acting on an object. It says that the rate of change of linear momentum of an object is directly proportional to the external applied force. It is given by :

F=\dfrac{\Delta p}{\Delta t}

Since, p = mv m is mass and v is velocity.

F=\dfrac{m(v-u)}{ t}\\\\F\times t=m(v-u)

So, if one knows only the constant resultant force acting on an object and the time during which this force acts, one can determine the change in momentum of the impulse.

5 0
3 years ago
An object, located 80.0 cm from a concave lens, forms an image 39.6 cm from the lens on the same side as the object. What is the
kramer

Answer:

-78.4 cm

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

Here we have

p = 80.0 cm

q = -39.6 cm (negative, because the image is on the same side as the object , so it is a virtual image)

Substituting, we find f:

\frac{1}{f}=\frac{1}{80.0 cm}+\frac{1}{-39.6 cm}=-0.0128 cm^{-1}

f=\frac{1}{0.0128 cm^{-1}}=-78.4 cm

5 0
4 years ago
IF YOU GET GOOD GRADES ON SCIENCE TESTS PLEASE HELP ME!!!!!
Ira Lisetskai [31]

Answer:

The answer is 100% A  THE ANSWER IS A!!!!

Explanation:

i have a 100 in all of my classes including science

8 0
3 years ago
Read 2 more answers
Convert 3402kgm/s to 20000Newtons
oee [108]

The 3,402 has units of kg-m/s.  That's momentum.  The 20,000 has units of Newtons.  That's force.  Momentum and force are different physical things, and you can't convert them from one to the other.

The best I can do for you is something like this:

Let's say you have a moving object with 3,402 kg-m/s of momentum, and you want to STOP it completely.  You want to stand in front of it and push back on it, hard enough and for long enough to CHANGE its momentum from 3,402 kg-m/s to zero.

Also ... there's a limit to how hard you can push.  The most force you can exert is 20,000 Newtons.

The amount you'll change its momentum is called the <u><em>impulse</em></u> you give it.  The quantity of impulse is (force) x (length of time you push on it).

So you need to keep pushing it back for (T seconds) long enough so that

(20,000 Newtons of force) x (T seconds) = 3,402 kg-m/s of momentum .

Divide each side of that equation by (20,000 Newtons). Then it says:

(T seconds) = (3,402 kg-m/s) / (20,000 Newtons)

<em>T = 0.1701 second</em>

And that's how you provide just enough impulse to stop the flying object ... push on it with 20,000 Newtons of force for exactly 0.1701 second, and it loses all its momentum and falls out of the air onto the ground at your feet.

This story is the closest I can come to anything that looks like "convert"ing momentum into force.

3 0
3 years ago
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