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sergeinik [125]
3 years ago
10

The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou

s atmospheric processes, including lightning. If research came out on Planet x in a distant solar system that had a electric field with strength 222 N/C and 0.6 the radius of the earth, what would be the excess charge on planet x
Physics
1 answer:
zlopas [31]3 years ago
4 0

Answer:

q = 3.6 10⁵  C

Explanation:

To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth

          r = 6 , 37 106 m

          E = k q / r²

          q = E r² / k

          q = \frac{100 \ (6.37 \ 10^6)^2}{9 \ 10^9}

          q = 4.5 10⁵ C

Now let's calculate the charge on the planet with E = 222 N / c and radius

           r = 0.6 r_ Earth

           r = 0.6 6.37 10⁶ = 3.822 10⁶ m

           E = k q / r²

            q = E r² / k

            q = \frac{222 (3.822 \ 10^6)^2}{ 9 \ 10^9}

            q = 3.6 10⁵  C

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Help me answer this question and get yourself some points!
Marrrta [24]

Answer:

A×B=C×D

500×0.5=250×X

250=250×X

X=250/250=1

X=1 m

Explanation:

note: if the force plus two, the distance will be half.

4 0
3 years ago
Uranium-235 decays to thorium-231 with a half-life of 700 million years. When a rock was formed, it contained 6400 million urani
Dahasolnce [82]

Answer:

proof in explanation

Explanation:

First, we will calculate the number of half-lives:

n = \frac{t}{t_{1/2}}

where,

n = no. of half-lives = ?

t = total time passed = 2100 million years

t_{1/2} = half-life = 700 million years

Therefore,

n = \frac{2100\ million\ years}{700\ million\ years}\\\\n = 3

Now, we will calculate the number of uranium nuclei left (n_u):

n_u = \frac{1}{2^{n} }(total\ nuclei)\\\\n_u = \frac{1}{2^{3} }(6400\ million)\\\\n_u = \frac{1}{8}(6400\ million)\\\\n_u =  800\ million

and the rest of the uranium nuclei will become thorium nuclei (u_{th})

n_{th} = total\ nuclei - n_u\\n_{th} = 6400\ million-800\ million\\n_{th} = 5600\ million

dividing both:

\frac{n_{th}}{n_u}=\frac{5600\ million}{800\ million} \\\\n_{th} = 7n_u

<u>Hence, it is proven that after 2100 million years there are seven times more thorium nuclei than uranium nuclei in the rock.</u>

6 0
3 years ago
In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a fur
Margaret [11]

Answer:

D = -4/7 = - 0.57

C = 17/7 = 2.43

Explanation:

We have the following two equations:

3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)

First, we isolate C from equation (2):

2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)

using this value of C from equation (3) in equation (1):

3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}

<u>D = - 0.57</u>

Put this value in equation (3), we get:

C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C =  \frac{17}{7}\\

<u>C = 2.43</u>

5 0
3 years ago
The actual mechanical advantage of any machine is its__________divided by its________________
zaharov [31]

Answer: Load divided by it effort

Explanation:

Mechanical advantage of any machine is its load divided by its effort

4 0
3 years ago
On a topographic map, contour lines create a group of concentric, closed loops. Which of the following features could this indic
posledela

Answer:

b. Hill top

Explanation:

On a topographic map, the closed circles are meant to represent a hill. So if the contour lines are creating a group of concentric closed loops then it must be an indication of a hill.

3 0
3 years ago
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