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alexira [117]
3 years ago
13

Consider a spherical Gaussian surface and three charges: q1 = 1.60 μC , q2 = -2.61 μC , and q3 = 3.67 μC . Find the electric flu

x through the Gaussian surface if it completely encloses (a) only charges q1 and q2, (b) only charges q2 and q3, and (c) all three charges.

Physics
1 answer:
Olenka [21]3 years ago
4 0

Answer:

Explanation:

<h3>Guass Law: Also known as  "Gauss's flux theorem" is the total of the electric flux "φ" out of a closed surface is equal to the charge "Q" enclosed divided by the permittivity εο. Solution is attached.</h3>

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5. In 1947 Bob Feller, a pitcher for the Cleveland Indians, threw a baseball across the
Darina [25.2K]

The maximum height reached by the ball is 99.2 m

Explanation:

When the ball is thrown straight up, it follows a free fall motion, which is a uniformly accelerated motion with constant acceleration (g=9.8 m/s^2 towards the ground). Therefore, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

In this problem, we have:

u = 44.1 m/s is the initial vertical velocity of the ball

v = 0 is the final velocity when the ball reaches the maximum height

s is the maximum height

a=-g=-9.8 m/s^2 is the acceleration of gravity (downward, so negative)

Solving for s, we find the maximum height reached by the ball:

s=-\frac{u^2}{2a}=-\frac{44.1^2}{2(-9.8)}=99.2 m

Learn more about free fall:

brainly.com/question/1748290

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3 0
4 years ago
On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-f
aliya0001 [1]

Answer:

15.3 s and 332 m

Explanation:

With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon

    gm = 1/6 ge

    gm = 1/6  9.8 m/s² = 1.63 m/s²

We calculate the range

    R = Vo² sin 2θ  / g

    R = 25² sin (2 30) / 1.63

    R= 332 m

We will calculate the time of flight,

   Y = Voy t – ½ g t2  

   Voy = Vo sin θ

When the ball reaches the end point has the same initial  height Y=0

0 = Vo sin  t – ½  g t2

0 = 25 sin (30)  t – ½ 1.63 t2

0= 12.5 t –  0.815 t2

We solve the equation

0= t ( 12.5 -0.815 t)

 t=0 s

t= 15.3 s

The value of zero corresponds to the departure point and the flight time is 15.3 s

Let's calculate the reach on earth

R2 = 25² sin (2 30) / 9.8

R2 = 55.2 m

R/R2 = 332/55.2

R/R2 = 6

Therefore the ball travels a distance six times greater on the moon than on Earth

5 0
3 years ago
Which of Newton's laws accounts for the following statement? "Negative acceleration is proportional to applied braking force." f
Vinil7 [7]
I believe it is Newtons Second Law! :)
7 0
3 years ago
Read 2 more answers
What happens to the current in a circuit if the resistance triples? If the voltage triples?
Sauron [17]
(Example 1 ) 
<span>If the Voltage that furnishes the current is an ideal (no internal resistance) Voltage source. Then; </span>

<span>V/R = i </span>
<span>V/2R = i/2 If external resistance doubles, current reduced to 1/2 of original value </span>
<span>V/3R = i/3 If external resistance triples, current reduced to 1/3 of original value </span>

<span>(Example 2) </span>
<span>But if the Voltage that furnishes the current is a practical [contains an internal resistance (Ri)] Voltage source. Then the current is a function of the Voltage source`s internal resistance, which does not double nor triple, plus the external resistance which is being doubled and tripled. </span>

<span>V/(R + Ri) = i </span>
<span>V/(2R + Ri) = greater than i/2 but less than I. </span>
<span>V/(3R + Ri) = greater than i/3 but less than i/2</span>
7 0
4 years ago
An airplane is flying through a thundercloud at a height of 2000 m (This is a very dangerous thing to do because of updrafts, tu
Doss [256]

Answer:

400000\ \text{N/C}

Explanation:

q_1 = Charge at 3000 m = 40 C

q_2 = Charge at 1000 m = -40 C

r_1 = 3000 m

r_2 = 1000 m

k = Coulomb constant = 9\times10^9\ \text{Nm}^2/\text{C}^2

Electric field due to the charge at 3000 m

E_1=\dfrac{k|q_1|}{r_1^2}\\\Rightarrow E_1=\dfrac{9\times 10^9\times 40}{3000^2}\\\Rightarrow E_1=40000\ \text{N/C}

Electric field due to the charge at 1000 m

E_2=\dfrac{k|q_2|}{r_2^2}\\\Rightarrow E_2=\dfrac{9\times 10^9\times 40}{1000^2}\\\Rightarrow E_2=360000\ \text{N/C}

Electric field at the aircraft is E_1+E_2=40000+360000=400000\ \text{N/C}.

7 0
3 years ago
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