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djyliett [7]
3 years ago
13

4. Provide an example of an isolated energy system and explain how it could be changed to create an open energy system. You may

use the examples provided in the lesson or another example.
Physics
2 answers:
levacccp [35]3 years ago
5 0

Answer:

Explanation:

Isolated energy system does not allow the energy or matter transfer, whereas, Open energy system allows energy and matter to exchange. For example, Insulated container (Thermos bottle), if the insulated bottle is filled with hot tea and the container is closed then there will be no exchange of matter and energy but as we open up the top of container then the heat will escape to the surrounding and exchange of matter and energy will take place.

GarryVolchara [31]3 years ago
4 0
<span>Easy, take the top off your Thermos bottle filled with hot coffee. Assuming perfect insulation, that hot coffee is isolated from the environment; but when the top is opened the heat can now escape to that environment.

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.
</span>
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A piece of metal has a mass of 10g and a mass of 2cm
qwelly [4]

Answer:

so whats the  questain?!

Explanation:

idgi

3 0
3 years ago
A material you are testing conducts electricity but cannot be pulled into wires. It is most likely a _____.
notsponge [240]

Answer: The correct answer is "metalloid".

Explanation:

Metal is the material which can conduct electricity as it contains free electrons. It is good conductor of electricity. For example, copper, silver.

Metal can be drawn into wires. This property of material is called ductile.

Metal can be beaten into sheets. This property of material is called malleable.

Non metal is the material which cannot conduct electricity as it does not contain free electrons. They are poor conductor of electricity. For example, oxygen.

Metalloid: It has properties of both metals and non metal. It is electrical conductor. For example, semiconductor- silicon and germanium. But they cannot be pulled into wires.

Therefore, a material you are testing conducts electricity but cannot be pulled into wires. It is most likely a metalloid.

6 0
3 years ago
A ball is thrown downward with an initial speed of 7 m / s. The ball's velocity after 3 seconds is m / s. (g = -9.8m / s²)
yan [13]

Answer:

-36.4 m/s

Explanation:

final velocity= initial velocity + acceleration x time

7 + (-9.8)(3)= -36.4 m/s

5 0
3 years ago
In a simple machine, the energy input is 120 J. If the efficiency of the machine is 80%, calculate the energy output
pashok25 [27]

Answer:

<h2>96 Joules</h2>

Explanation:

We know that efficiency is the ratio of output power by input power. i.e. Efficiency describes the quality of machine or system how good it is.

Solution,

Energy input of system = 120 J

Efficiency = 80% = \frac{80}{100}  = 0.8

Now,

According to definition,

Efficiency = \frac{output}{input}

Cross multiplication:

output \:  =  \: 0.8 \times 120

Calculate the product

output \:  = 96 \: joules

Hope this helps...

Good luck on your assignment...

4 0
3 years ago
You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C
jarptica [38.1K]

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

4 0
4 years ago
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