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Law Incorporation [45]
2 years ago
9

A can of soft drink at room temperature is put into the refrigerator so that it will cool. Would you model the can of soft drink

as a closed system or as an open system?
Physics
1 answer:
Gennadij [26K]2 years ago
7 0

Answer:

A can of soft drink being transferred from the room temperature to a refrigerator is a closed system.

Explanation:

When a can of soft drink is transferred to a refrigerator from room temperature then only energy transfer can occur i.e. the heat of soft drink is absorbed by the refrigerator but the quantity of mass packed in the can does not gets altered.

From the thermodynamic point of view a closed system is one in which the mass content of the system does not changes only the energy transfer can take place through the boundary of the system.

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A microprocessor scans the status of an output I/O device every 20 ms. This is accomplished by means of a timer alerting the pro
Lerok [7]

Answer:

0.0000045 s

Explanation:

f = Frequency = 8 MHz

Clock cycle is given by

\dfrac{1}{f}=\dfrac{1}{8\times 10^6}=1.25\times 10^{-7}\ s

Time taken for 12 clock cycles

12\times 1.25\times 10^{-7}=0.0000015\ s

Time taken per instruction is 0.0000015 s

In reading and displaying information it requires 3 processes

1 for reading, 1 for searching and 1 for displaying.

3\times 0.0000015=0.0000045\ s

Time taken is 0.0000045 s

6 0
1 year ago
A bucket of water experiencing a gravitational force of 525 N is pulled up from a water well. The net force in the y-direction i
lukranit [14]

Answer:

6n!!!!!!!!!!!!!!!!!!

Explanation:

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8 0
1 year ago
Which of the following forces exists between objects even in the absence of direct physical contact
den301095 [7]

Answer: TRUST ME I GOT IT WRONG the answer is B

Explanation:

3 0
2 years ago
Read 2 more answers
The earth's magnetic field, with a magnetic dipole moment of 8.0×1022Am2, is generated by currents within the molten iron of the
Fed [463]

To solve this problem it is necessary to apply the concepts related to the magnetic dipole moment in terms of the current and the surface area, as well as the current density, as a function of the current over the area.

Part A) By definition we know that magnetic dipole moment is

m = IS

Where,

I = Current

S = Area

m = IA \rightarrow m= I( \pi r^2)

Replacing with our values we have that,

8*10^{22} = I \pi(\frac{3000*10^3}{2})^2

Re-arrange to find I,

I = 1.1317*10^{10} A

Part B) To find the Current density we need to find the cross sectional area of the Wire:

A = \pi r^2 \\A = \pi (\frac{1000*10^3}{2})^2\\A = 7.854*10^{11}m^2

Finally the current density is simply J

J = \frac{I}{A}\\J = \frac{1.1317*10^{10}}{7.854*10^{11}}\\J = 0.0144A/m^2

PART C) Finally to make the comparison with the given values we have to cross-sectional area would be

A = \pi (10-3)^2 \\A = 49\pi

Therefore the current density would be

J = \frac{I}{A}\\J = \frac{30A}{49\pi}\\J = 9.549*10^5 A/m^2

Comparing the two values we can see that the 2mm wire has a higher current density.

4 0
1 year ago
A small child gives a plastic frog a big push at the bottom of a slippery 2.0 meter long, 1.0 meter high ramp, starting it with
valentinak56 [21]
Refer to the diagram shown below.

Because the ramp is slippery, ignore dynamic friction.
Let m =  the mass of the frog.
g = 9.8 m/s²

The KE (kinetic energy) at the bottom of the ramp is
KE₁ = (1/2)*(m kg)*(5 m/s)² = 12.5 m J

Let v =  the velocity at the top of the ramp.
The KE at the top of the ramp is
KE₂ = (1/2)*m*v²= 0.5 mv² J
The PE (potential energy) at the top of the ramp relative to the bottom is
PE₂ = (m kg)*(9.8 m/s²)*(1 m) = 9.8m J

Conservation of energy requires that
KE₁ = KE₂ + PE₂
12.5m = 0.5mv² + 9.8m
0.5v² = 2.7
v = 2.324 m/s

Answer: 2.324 m/s

7 0
2 years ago
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