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maria [59]
2 years ago
5

Determine the values of mm and nn when the following average distance from the Sun to the Earth is written in scientific notatio

n: 150,000,000,000 mm.
Physics
1 answer:
viktelen [127]2 years ago
5 0

Answer:

m × 10^n = 1.5 × 10^11 mm

m = 1.5

n = 11

Explanation:

Scientific notation is written in the form;

m × 10^n

Where m = a number between 1 and 10

And n is the power or exponent.

To convert

150,000,000,000 mm

we need to know the number of movement of the decimal before we reach a number between 1 and 10.

There are 11 movement to reach 1.5 and since the movement is to the left then the power carries a positive sign. So, the scientific notation is given as;

m × 10^n = 1.5 × 10^11 mm

m = 1.5

n = 11

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Given the indices of refraction n1 and n2 of material 1 and material 2, respectively, rank these scenarios on the basis of the p
lisov135 [29]

Answer:

c>d>f=a>b>e

Explanation:

When a pair of medial has greater difference between the their individual refractive indices with respect to vacuum then it has a greater deviation between the refracted ray and the incident ray.

According to the Snell's law:

\rm refractive\ index\ (n)=\frac{speed\ of\ light\ in\ the\ incident\ medium}{speed\ of\ light\ in\ the\ refracted\ medium}

a)

n_1-n_2=1.33-1.00\\=0.33

b)

n_2-n_1=1.46-1.33

=0.23

c)

n_2-n_1=2.42-1.33\\=1.09

d)

n_2-n_1=1.46-1.00\\=0.46

e)

n_1-n_2=1.50-1.33\\=0.17

f)

n_2-n_1=1.33-1.00\\=0.33

c>d>f=a>b>e

5 0
2 years ago
In 1923, the United States Army (there was no U.S. Air Force at that time) set a record for in-flight refueling of airplanes. Us
dybincka [34]

Answer:

1.95m/s

Explanation:

Please view the attached file for the detailed solution.

The following were the conversion factors used in order to express all quatities in SI units:

1 gallon=0.00378541m^3\\1 inch=0.0254m\\1 minute=60s

6 0
2 years ago
A runner runs 4875 ft in 6.85 minutes. what is the runners average speed in miles per hour?
yanalaym [24]

The average speed can be easily calculated by taking the ratio of distance and time. That is:

average speed = distance / time

 

so calculating:

average speed = 4875 ft / 6.85 minutes

<span>average speed = 711.68 ft / min</span>

8 0
2 years ago
Read 2 more answers
You should have observed that there are some frequencies where the output is stronger than the input. Discuss how that is even p
nydimaria [60]

Answer:

w = √ 1 / CL

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

Explanation:

This problem refers to electrical circuits, the circuits where this phenomenon occurs are series RLC circuits, where the resistor, the capacitor and the inductance are placed in series.

In these circuits the impedance is

             X = √ (R² +  (X_{C} -X_{L})² )

where Xc and XL is the capacitive and inductive impedance, respectively

            X_{C} = 1 / wC

           X_{L} = wL

From this expression we can see that for the resonance frequency

           X_{C} = X_{L}

the impedance of the circuit is minimal, therefore the current and voltage are maximum and an increase in signal intensity is observed.

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

               V = IR

Since the contribution of the two other components is canceled, this occurs for

                X_{C} = X_{L}

                1 / wC = w L

                w = √ 1 / CL

6 0
2 years ago
A 5.00-pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to
Ne4ueva [31]

Answer:

a) r=4.24cm d=1 cm

b) Q=5x10^{-10} C

Explanation:

The capacitance depends only of the geometry of the capacitor so to design in this case knowing the Voltage and the electric field

V=1.00x10^{2}v\\E=1.00x10^{4} \frac{N}{C}

V=E*d\\d=\frac{V}{E}\\d=\frac{1.0x10^{2}}{1.0x10^{4}}\\d=0.01m

The distance must be the separation the r distance can be find also using

C=\frac{Q}{V_{ab}}

But now don't know the charge these plates can hold yet so

a).

d=0.01m

C=E_{o}*\frac{A}{d}\\A=\frac{C*d}{E_{o}}

A=\frac{5pF*0.01m}{8.85x10^{-12}\frac{F}{m}}\\A=5.69x10^{-3}m^{2}

A=\pi *r^{2}\\r=\sqrt{\frac{A}{r}}\\r=\sqrt{\frac{5.64x10^{-3}m^{2} }{\pi } }  \\r=42.55x^{-3}m

b).

C=\frac{Q}{V_{ab}}

Q=C*V\\Q=5x10^{-12} F*1x10^{2}\\Q=5x10^{-10}C

8 0
2 years ago
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