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PtichkaEL [24]
3 years ago
9

How many electrons make up a charge of 3.5kC?

Physics
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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Light travels 300,000,000m/sec, and one year has approximately 32,000,000 seconds. A light year is the distance light travels in
dem82 [27]
A) 3 x 10 ^ 8
b) 3 x 10 ^ 5
c) 3.2 x 10 ^ 7
d) 9.6 x 10 ^ 15 m
e) 9.6 x 10 ^ 17 cm
3 0
3 years ago
Which category of galaxy does not have a distinctive shape?
mylen [45]
The category of galaxy which does not have a distinctive shape is D. an irregular galaxy.
A spiral galaxy has a spiral shape, an elliptical galaxy has an elliptical shape, and a barred-spiral galaxy has a barred-spiral shape. The only galaxy type which does not have a constant shape is an irregular galaxy.
5 0
3 years ago
Read 2 more answers
Please help! Will give brainliest. 10 points. Show work!
Natasha_Volkova [10]

Answer:

421.83 m.

Explanation:

The following data were obtained from the question:

Height (h) = 396.9 m

Initial velocity (u) = 46.87 m/s

Horizontal distance (s) =...?

First, we shall determine the time taken for the ball to get to the ground.

This can be calculated by doing the following:

t = √(2h/g)

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 396.9 m

Time (t) =.?

t = √(2h/g)

t = √(2 x 396.9 / 9.8)

t = √81

t = 9 secs.

Therefore, it took 9 secs fir the ball to get to the ground.

Finally, we shall determine the horizontal distance travelled by the ball as illustrated below:

Time (t) = 9 secs.

Initial velocity (u) = 46.87 m/s

Horizontal distance (s) =...?

s = ut

s = 46.87 x 9

s = 421.83 m

Therefore, the horizontal distance travelled by the ball is 421.83 m

5 0
2 years ago
A drag racer starts from the rest and accelerates at 7.4m/s^2.how far will he travel in 2.0 seconds?
Vesna [10]

Answer:

14.8 m

Explanation:

S= ut + \frac{1}{2}at^{2}

where u = initial velocity

S= (0 \frac{m}{s})(2s) + \frac{1}{2}(7.4\frac{m}{s^{2} })(2s^{2})

S=  \frac{1}{2}(7.4\frac{m}{s^{2} })(2s^{2})

S=14.8 m

8 0
3 years ago
Could I please get some help on this question I don’t understand .
Oksana_A [137]

Answer:

12.5 m/s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Height (h) = 8 m

Final velocity (v) at 8 m above the lowest point =?

NOTE: Acceleration due to gravity (g) = 9.8 m/s²

The velocity of the roller coaster at 8 m above the lowest point can be obtained as follow:

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 8)

v² = 0 + 156.8

v² = 156.8

Take the square root of both side

v = √156.8

v = 12.5 m/s

Therefore, the velocity of the roller coaster at 8 m above the lowest point is 12.5 m/s.

5 0
3 years ago
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