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KIM [24]
3 years ago
11

In the lesson a thermos is presented as an example of an isolated energy system. How could you change the thermos into an open e

nergy system?
Physics
2 answers:
iVinArrow [24]3 years ago
7 0
If you want to change the thermos into an open energy system, you have to remove the lid. Once the lid is removed, the energy is no longer contained inside the thermos bottle. From the bottle, the energy dissipates to the environment.
Vitek1552 [10]3 years ago
5 0

Answer:

The lid of the thermos must be removed.

Explanation:

The definition of an open system is one that can exchange heat and mass with the surrounding medium. A closed thermos cannot exchange mass or temperature with the medium. By removing the lid, it has the possibility of exchanging mass and energy in the form of heat with the medium in which it is found.

Have a nice day!

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1) A rock is dropped from a cliff with a height of 200 m. With what velocity will the rock hit the ground
Komok [63]

A rock is dropped from a 200 m high cliff. How long does it take to fall (a) the first 100 m and (b) the last 50 m?

The basic equation you want is:

s=at22

Solving for t:

t=2sa−−−√

We’ll assume a=9.8 , so 2a−−√=14.9−−−√≈0.4518

So, for (a)s=100 , so t=0.4518100−−−√=4.518

The total time is 0.4518200−−−√≈6.389

The time to fall 150 m is 0.4518150−−−√≈5.533

So the time to fall the last 50 m is 6.389 - 5.533 = 0.856 seconds

(

6 0
2 years ago
Read 2 more answers
A 3.0 kg mass is released from rest at point A. The mass slides along the curved surface to point B in 6.0 seconds. Point B is 2
timurjin [86]

Answer:

option d) -9 J

Explanation:

Given:

Mass, m = 3.0 kg

time, t = 6.0 seconds

Velocity of mass, v = 2.0 m/s

height, h = 2 m

Now, using the concept of work-Energy theorem

we have

Net work done = change in kinetic energy

or

Work done by gravity + work done by the friction = Final kinetic energy - Initial kinetic energy

mgh +W_f = \frac{1}{2}mv^2-0

on substituting the values in the above equation, we get

3 × 9.8 × 2 + W_f = \frac{1}{2}\times 3\times2^2

or

58.8 + W_f = 6

or

W_f = -52.8 J

here negative sign depicts that the work is done against the motion of the mass

also,

Power = (Work done)/time

or

Power = -52.8/6 = -8.8 W ≈ 9 J

Hence, option d) -9 J is correct

5 0
3 years ago
Unit Test Review
PolarNik [594]

Answer:

option B...

they represent different concept...

i hope this will answer your question

4 0
1 year ago
Identify some sports or activities you can do to improve your overall health and explain how physical activity can improve your
Alexxx [7]

Answer:

soccer

Explanation:

soccer can inprove your health because you use all parts of your body playing it. You use your legs because you are running and you use your arms for doing throw ins. Soccer is also good for you because you are not playing for too long. Usually one soccer game is only 1 hour.

Hope this helps

3 0
3 years ago
The Mars Curiosity rover was required to land on the surface of Mars with a velocity of 1 m/s. Given the mass of the landing veh
Aliun [14]

Answer:

The value is      A   = 39315 \  m^2

Explanation:

From the question we are told that

    The velocity which the rover is suppose to land with is  v  =  1 \ m/s

    The  mass of the rover and the parachute is  m  =  2270 \ kg

     The  drag coefficient is  C__{D}}  =  0.5

      The atmospheric density of Earth  is  \rho =  1.2 \  kg/m^3

     The acceleration due to gravity in Mars is  g_m  =  3.689 \  m/s^2

     

Generally the Mars  atmosphere density is mathematically represented as

          \rho_m  =  0.71 *  \rho

=>        \rho_m  =  0.71 *  1.2

=>        \rho_m  = 0.852 \  kg/m^3

Generally the drag force on the rover and the parachute  is mathematically represented as

          F__{D}} =  m  *  g_{m}

=>       F__{D}} =  2270   *  3.689  

=>       F__{D}} =  8374 \ N  

Gnerally this drag force is mathematically represented as

         F__{D}} =   C__{D}} *  A *  \frac{\rho_m * v^2 }{2}

Here A is the frontal area

So  

         A   =  \frac{2 *  F__D }{ C__D}  *  \rho_m  * v^2   }

=>       A   =  \frac{2 * 8374 }{ 0.5 *  0.852    *  1 ^2   }

=>       A   = 39315 \  m^2

8 0
3 years ago
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