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Mnenie [13.5K]
2 years ago
12

What does the letter I represent?

Physics
1 answer:
irina [24]2 years ago
6 0
The letter i is used to signify that a number is an imaginary number. It stand for the square root of negative one.
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). with the input voltage range set at +/- 500mv, what is the smallest difference in voltage that can be resolved? show your cal
dalvyx [7]
The smallest difference in voltage that can be resolved is referred to as the resolution. The resolution can be calculated with the following formula:
resolution=voltage range / digital range
The voltage range in our case is from -500mV to 500mV, which gives 1000mV.
The digital range on the other hand is 2^(number of bits).
It depends on what type of bit board we are using. If the ADC we are using is a 16 bit board, then 2^16=<span>65536.
So, the resolution is:
resolution=1000mV/</span><span>65536=0.015 mV</span>
4 0
3 years ago
two charges + 2.6 uc and -5.4uc experience an attractive force of 6.5mN. What is the separation between the charges?
djyliett [7]

Answer:

...

Explanation:

....

7 0
3 years ago
A car is traveling at an average speed of 70 m/s. How many km would the car travel in 6.5 hrs. ?
docker41 [41]

Answer:

<h2>38,769.23 miles</h2>

Explanation:

given:

A car is traveling at an average speed of 70 m/s.

find:

How many km would the car travel in 6.5 hrs. ?

solution:

distance = velocity over time

let velocity = 70 m/s

           time = 6.5 hrs.

convert velocity 70 m/s into m/h for consistency of units.

<u>70 mi. </u> x   <u>3600 sec.</u>  =  252,000 mi/hour

  sec.          1 hr.

now plugin values into the formula d = v/t

d = <u>252,000 miles/hour </u>

              6.5 hours

d = 38,769.23 miles

therefore, the distance travelled in 6.5 hours with a speed of 70 m/s is 38,769.23 miles

5 0
3 years ago
"The drawing shows three layers of different materials, with air above and below the layers. The interfaces between the layers a
My name is Ann [436]

Answer:

Angle of incidence that entered material b= 63.1°

Angle of incidence between a and b = 55.9°

Explanation: Using the formular:

n1sintheta1= n2sintheta2

The light ray which enters material B will be

1.4Sin72.8° = 1.5Sin theta

1.3373= 1.5Sintheta

sintheta = 1.3373/1.5

Sin^-1 0.8916 = Theta

63.1 = theta

When the ray hits interface with material a

1.5Sin63.1 = 1.3 Sin theta

1.3374 = 1.3Sin theta

Sintheta= 1.3374/1.3

Sin theta = 1.0877

There will be total reflection off the boundary b c because sin theta exceeded 1 in value.

The equation should be

1.4sin63.1 = 1.4 sin theta

Sin theta=72.8°

When the ray hits air-c boundary:

1.4sin72.8=1.00sin theta

Sin theta=1.3374/1 =1.3374

There is total reflection.

In material a,the ray will:

1.3sin72.8° = 1.00sin theta

There will be total reflection when the ray hits a-b boundary.

1.3sin72.8= 1.5sintheta

Sin theta= 1.2419/ 1.5

Sin theta =0.8279

Theta= Sin^-10.8279= 55.88°

When ray hits c-air boundary

1.4sin63.1= 1.00sintheta

1.2485= sin theta = Toal reflection.

Therefore when the ray of light pass through the layers of material a, b and c the boundary with air on top and bottom will be total reflection.

7 0
3 years ago
A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800
Arada [10]

Explanation:

The given data is as follows.

            m = 5000 kg,            h = 800 km = 0.8 \times 10^{6} m

    R_{e} = 6.37 \times 10^{6} m,    r = R + h = 7.17 \times 10^{6} m

   M_{e} = 5.98 \times 10^{24} kg,   G = 6.67 \times 10^{-11} Nm^{2}/kg^{2}

As we know that,

              \frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}

                      v = \sqrt{\frac{GM_{e}}{r^{2}}}

And, it is known that formula to calculate angular velocity is as follows.

               \omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}

                            v = \sqrt{\frac{GM_{e}}{r^{3}}}

                               = \sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}

                                = 1.0402 \times 10^{-3} rad/s

Thus, we can conclude that speed of the satellite is 1.0402 \times 10^{-3} rad/s.

6 0
3 years ago
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