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3 years ago

A typical lightning bolt has about 10.0 c of charge. how many excess electrons are in a typical lightning bolt?

2 answers:

5 0

The charge of an electron is -1.602 x 10^-19 Coulombs. If an average lightning bolt carries around 10.0 Coulombs of charge, we can simply divide this by the charge of an electron to estimate the total number of excess electrons:

(10.0 C) / (1.602 x 10^-19 C/electron) = 6.24 x 10^19 electrons
So there are around 6.24 x 10^19 excess electrons in a typical lightning bolt.

lubasha [3.4K]3 years ago

4 0

Answer: $6.24\times 10^{19} electrons$ are present in the lightening bolt.

Explanation:

Total charge = -10.0 C

Charge on 1 electron = $-1.602\times 10^{-19} C$

Number of electrons = n

Total charge = n × charge on 1 electron

$-10.0 C=n\times -1.602\times 10^{-19} C$

$n=\frac{-10.0 C}{-1.602\times 10^{-19} C}=6.24\times 10^{19} electrons$

$6.24\times 10^{19} electrons$ are present in the lightening bolt.

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A grapefruit falls from a tree and hits the ground 0.76 s later.

Tomtit [17]

Answer:

2.8 m 7.4 m/s

Explanation:

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3 years ago

A bowl contains 4 red balls and 2 green balls. Two balls are selected simultanesouly and at random. What is the conditional prob

AlladinOne [14]

Answer:0.6

Explanation:

Given

Bowl contains 4 red and 2 Green balls

Probability of selecting two red balls given atleast one of the ball is red + Probability of selecting two red balls given that no ball is red =1

Probability of selecting two red balls given that no ball is red $=\frac{Probability \ of\ selecting\ two\ red\ balls\ out\ of\ 4\ balls}{probability\ of\ selecting\ 2\ balls\ out\ of\ 6 balls}$

$P=\frac{^{4}C_2}{^{6}C_2}$

$P=\frac{2}{5}$

Required Probability $=1-\frac{2}{5}=\frac{3}{5}$

7 0

4 years ago

In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal.

HACTEHA [7]

A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is $41.9^{\circ}$

Explanation:

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

$s=u_y t + \frac{1}{2}at^2$ (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

$u_y = u sin \theta$ is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

$\theta=40.0^{\circ}$ is the angle of projection

$u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s$

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (downward)

Substituting the numbers, we get

$-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0$

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

$d=u_x t$

where

$u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s$ is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

$d=(9.19)(1.778)=16.34 m$

In this second case,

$\theta=42.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.1t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s$

So, the range of the shot is

$d=u_x t = (8.85)(1.851)=16.38 m$

In this third case,

$\theta=45^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.5t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s$

So, the range of the shot is

$d=u_x t = (8.49)(1.925)=16.34$ m

In this 4th case,

$\theta=47.5^{\circ}$

So the vertical velocity is

$u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s$

So the equation for the vertical motion becomes

$4.9t^2-8.8t-1.80=0$

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

$u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s$

So, the range of the shot is

$d=u_x t = (8.11)(1.981)=16.07 m$

From the previous parts, we see that the maximum range is obtained when the angle of releases is $\theta=42.5^{\circ}$ .

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

$s-u sin \theta t + \frac{1}{2}gt^2=0$

The solutions of this quadratic equation are

$t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}$

This is the time of flight: so, the horizontal range is

$d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta$

It can be found that the maximum of this function is obtained when the angle is

$\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})$

Therefore in this problem, the angle which leads to the maximum range is

$\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0

4 years ago

How many moles are in 73.4 grams of Phosphorus?

mylen [45]

23.2

hope this helped

4 0

3 years ago

Gold in its pure form is too soft to be used for most jewelry. Therefore, the gold is mixed with other metals to produce an allo

Novay_Z [31]

Answer:

Explanation:18kt alloy contains

i) 75% of gold

rhogold=19.3g/cm^3

=75/100×19.3

=14.475g/cm^3

ii) 16% of silver

rhosilver=10.5g/cm^3

=16/100×10.5

=1.68g/cm^3

iii) 9% of copper

rhocopper =8.90g/cm^3

=9/100×8.9

=0.801g/cm^3

Overall density of 18kt gold

=(0.801+1.68+14.475)g/cm^3

=16.956g/cm^3

=17g/cm^3 to 3s.f

6 0

3 years ago