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anygoal [31]
3 years ago
8

If the solar system shrank so that the sun were located just one centimeter from Earth, about how far away could you find the ne

xt nearest star?
Physics
1 answer:
DedPeter [7]3 years ago
8 0
E                  S                                                               *

The "E" represents Earth, "S" represent Sun, and the "*" represents the nearest star(which is Proxima Centauri).

The main thing to worry about here is units, so ill label everything out.
D'e,s'(Distance between earth and sun) = .<span>00001581 light years
D'e,*'(Distance between earth and Proxima) = </span><span>4.243 light years

Now this is where it gets fun, we need to put all the light years into centimeters.(theres alot)
In one light year, there are </span>9.461 * 10^17 centimeters.(the * in this case means multiplication) or 946,100,000,000,000,000 centimeters.

To convert we multiply the light years we found by the big number.
D'e,s'(Distance between earth and sun) = 1.496 * 10^13 centimeters<span>
D'e,*'(Distance between earth and Proxima) = </span><span>4.014 * 10^18 centimeters
</span>
Now we scale things down, we treat 1.496 * 10^13 centimeters as a SINGLE centimeter, because that's the distance between the earth and the sun. So all we have to do is divide (4.014 * 10^18 ) by (<span>1.496 * 10^13 ).
Why? because that how proportions work.

As a result, you get a mere 268335.7 centimeters.

To put that into perspective, that's only about 1.7 miles

A lot of my numbers came from google, so they are estimations and are not perfect, but its hard to be on really large scales.</span>
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Water flows at 10 m/s through a pipe with radius 0.025 m. The pipe goes up to the second floor of the building, 2.5 m higher, an
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Answer: from the information given, the velocity of the water will decrease but the pipe size will remain the same.

This can be proved with bernoulli's equation.

Explanation: careful analysis of the system using bernoulli's equation of flow is shown in the image attached

5 0
3 years ago
4.77 Augment the rectifier circuit of Problem 4.70 with a capacitor chosen to provide a peak-to-peak ripple voltage of (i) 10% o
goblinko [34]

The question incomplete! The complete question along with answer and explanation is provided below.

Question:

Augment the rectifier circuit of Problem 4.68 with a  capacitor chosen to provide a peak-to-peak ripple voltage of  (i) 10% of the peak output and (ii) 1% of the peak output. In  each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(c) What is the average diode current?

(d) What is the peak diode current?

Problem 4.68:

A half-wave rectifier circuit with a 1-kΩ load operates from a 120-V (rms) 60-Hz household supply through  a 10-to-1 step-down transformer. It uses a silicon diode  that can be modeled to have a 0.7-V drop for any current.

Given Information:

Input voltage = 120 Vrms

10 to 1 step-down transformer

Voltage drop at diode = 0.7 V

Load resistance = R = 1 kΩ

Required Information:

 (i) 10% of the peak output and (ii) 1% of the peak output. In  each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(c) What is the average diode current?

(d) What is the peak diode current?

Answer:

Case (i)

Vavg = 15.45 V

Conduction of diode = 7.11 %

Iavg = 0.232 A

Ip = 0.449 A

Case (ii)

Vavg = 16.18 V

Conduction of diode = 2.25 %

Iavg = 0.735 A

Ip = 1.453 A

Explanation:

Voltage at the secondary side of the transformer is

Vrms = Vpri/turn ratio

Vrms = 120/10 = 12 V

The relation between rms voltage and peak voltage is

Vp = Vrms/√2

Vp = 12√2 = 16.97 V

Vd = 0.7 V

First we will calculate all the required parameters for the 10% ripple voltage and then for 1% ripple voltage.

case (i) 10% of the peak output:

(a) What average output voltage results?

Average output voltage = Vavg = Vp - Vd - 0.5Vr

Where Vp is the peak output voltage Vd is the voltage drop of diode and Vr is the ripple voltage which is given as a percentage of Vp

Vavg = Vp - Vd - 0.5Vr

Vavg = 16.97 - 0.7 - 0.5[0.1(16.97 - 0.7)]

Vavg = 15.45 V

(b) What fraction of the cycle does the diode conduct?

ω = √2Vr/Vp - Vd

ω = √2*0.1(Vp-Vd)/Vp - Vd

ω = √2*0.1(16.97-0.7)/16.97 - 0.7

ω = 0.447 rad

Conduction of diode = (ω/2π)*100

Conduction of diode = (0.447/2π)*100

Conduction of diode = 7.11 %

(c) What is the average diode current?

Average current = Iavg = Vavg/R[ 1 + π( √2(Vp - Vd)/0.1(Vp-Vd))]

Average current = Iavg = 15.45/1000[ 1 + π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Average current = Iavg = 0.232 A

(d) What is the peak diode current?

Peak current = Ip = Vavg/R[ 1 + 2π( √2(Vp - Vd)/0.1(Vp-Vd))]

Peak current = Ip = 15.45/1000[ 1 + 2π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Peak current = Ip = 0.449 A

case (ii) 1% of the peak output:

(a) What average output voltage results?

Vavg = 16.97 - 0.7 - 0.5[0.01(16.97 - 0.7)]

Vavg = 16.18 V

(b) What fraction of the cycle does the diode conduct?

ω = √2*0.01(Vp-Vd)/Vp - Vd

ω = √2*0.01(16.97-0.7)/16.97 - 0.7

ω = 0.1417 rad

Conduction of diode = (0.1417/2π)*100

Conduction of diode = 2.25 %

(c) What is the average diode current?

Average current = Iavg = 16.18/1000[ 1 + π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Average current = Iavg = 0.735 A

(d) What is the peak diode current?

Peak current = Ip = 16.18/1000[ 1 + 2π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Peak current = Ip = 1.453 A

3 0
3 years ago
For questions 1 - 4 complete the representations for the four patterns below. Provide the mathematical
larisa [96]

Answer:

Explanation:

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6 0
3 years ago
In an inelastic collision a 2.5 kg ball moving at 7.5 m/s is caught by a 70kg man while the man is standing on ice. What is the
MrRa [10]

The velocity of the ball and the man is 0.259 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, in an isolated system, the total momentum before and after the collision must be conserved. Therefore, for the ball-man system, we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v

where:

m_1 = 2.5 kg is the mass of the ball

u_1 = 7.5 m/s is the initial velocity of the ball

m_2 = 70 kg is the mass of the man

u_2 = 0 is the initial velocity of the man

v is the final velocity of the man and the ball after the collision

Re-arranging the equation and substituting the values, we find the final velocity:

v=\frac{m_1 u_1}{m_1+m_2}=\frac{(2.5)(7.5)}{2.5+70}=0.259 m/s

So, the man and the ball slides on the ice at 0.259 m/s.

Learn more about momentum:

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3 years ago
Which scenario is in excample of a scientific way to think
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Can you explain this a bit more I don’t quite understand
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