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3 years ago

How many electrons make up a charge of 3.5kC?

2 answers:

vladimir1956 [14]3 years ago

7 0

Well now, I don't know !
Let's figure it out.

The charge on one electron is 1.60 x 10⁻¹⁹ Coulomb .

To get 3,500 Coulombs of charge, you'd need to
go around and collect

   (3,500) / (1.60 x 10⁻¹⁹) electrons.

That's 2.19 x 10²² of the little fellas.

That's a big number. But what's REALLY amazing is the
mass and weight of that many electrons:

   Mass: 1.989 x 10⁻⁵ gram

   Weight: about 0.0000007 ounce !

That's not one electron. That's the whole <span>2.19 x 10²² uvvum
that it takes to hold 3,500 Coulombs of charge.</span>

ollegr [7]3 years ago

6 0

21.8452851 * 10^21 e. Which is 21.845.285.100.000.000.000.000 electrons. A lot.

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When a bow is stretched back and an arrow is shot, what type of energy conversion has occurred?

Lelechka [254]

D. mechanical to elastic

6 0

3 years ago

It is now 9:11 a.m. but when the bell rings at 9:12 a.m. Susie will be late for Mrs. Garner's U.S. History class for the 3rd tim

GaryK [48]

Answer:

3.1 m/s

Explanation:

The total distance she has to run is the addition of the three lengths:

47 + 63 + 76 = 186 meters.

She needs to cover it one minute (60 seconds). Therefore her speed must be:

186 m / 60 s = 3.1 m/s

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3 years ago

A force of 30 N is applied to an object with a mass of 15 kg. What is the resulting acceleration?

svet-max [94.6K]

Answer:

We know that Force = mass × acceleration

By substituting the values we get,

30 N = 15 kg × a (where a is acceleration)

Or we can write it as

15 kg × a = 30 N

Transposing 15 to RHS,

a = 30 ÷ 15 m/s²

Therefore, acceleration = 2 m/s²

pls give brainliest for the answer

5 0

3 years ago

Read 2 more answers

An object is located 50 cm from a converging lens having a focal length of 15 cm. Which of the following is true regarding the i

crimeas [40]

Answer:

It is real, inverted, and smaller than the object.

Explanation:

Let's start by using the lens equation to find the location of the image:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

where we have:

q = ? is the distance of the image from the lens

f = 15 cm is the focal length (positive for a converging lens)

p = 50 cm is the distance of the object from the lens

Solving the equation for q, we find

$\frac{1}{q}=\frac{1}{15 cm}-\frac{1}{50 cm}=0.047 cm^{-1}$

$q=\frac{1}{0.047 cm^{-1}}=+21.3 cm$

The sign of q is positive, so the image is real.

Now let's also write the magnification equation:

$h_i = - h_o \frac{q}{p}$

where

$h_i, h_o$ are the size of the image and of the object

By substituting p = 50 cm and q = 21.3 cm, we find

$h_i = - h_o \frac{21.3 cm}{50 cm}=-0.43 h_o$

So we notice that:

$|h_i| < |h_o|$ : this means that the image is smaller than the object

$h_i < 0$ : this means that the image is inverted

so, the correct option is:

It is real, inverted, and smaller than the object.

7 0

3 years ago

If a migrating loon flies at an average speed of 19 m/s, what is its average speed in mi/hr?

zlopas [31]

Unit conversion is a way of converting some common units into another without changing their real value. The average speed of the migrating loon flies is 45.5019 miles/hr.

<h3>What is Unit conversion?</h3>

Unit conversion is a way of converting some common units into another without changing their real value. for, example, 1 centimeter is equal to 10 mm, though the real measurement is still the same the units and numerical values have been changed.

Given that migrating loon flies at an average speed of 19 m/s. Now, since 1 meter is equal to (1/1609.34) miles and 1 second is equal to (1/3600) seconds. Therefore, we can write the speed as,

$\rm Speed = 19\ \dfrac{meter}{second}\\$

$=19 \times \dfrac{\frac{1}{1609.34}\rm\ miles}{\frac{1}{3600} \rm\ hours}$

= 19 × (3600/1609.34) miles/hr

= 42.5019 miles/ hr

Hence, the average speed of the migrating loon flies is 45.5019 miles/hr.

Learn more about Units conversion here:

brainly.com/question/4736731

#SPJ4

4 0

1 year ago