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Oxana [17]
2 years ago
8

Anybody help me, please? Describe one periodic trend that you learned about.

Physics
1 answer:
Korolek [52]2 years ago
8 0

Answer:

the renegade

Explanation: charklie dfamielo

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What kind of charging occurs when you slide your body across a plastic surface?
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<span>Charging by friction occurs, Electrons are transferred when one object rubs against another.

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3 years ago
You hold a ruler that has a charge on its tip 6 cm above a small piece of tissue paper to see if it can be picked up. The ruler
scZoUnD [109]

Answer:

q=1.4*10^{-9}C

Explanation:

Given data:

charge on ruler = -14μC

Mass of tissue is 5 g

To Know the minimum charge,  equate electrostatic force to weight  

we have F = W

so\frac{KQq}{r^2} =mg

putting all value in equation,

=\frac{9*10^9*(14*10^{-6})*q}{0.06^2} = 5* 10^{-3}*9.8

solving for q

q =\frac{5* 10^{-3}*9.8 *0.06^2}{9*10^9*(14*10^{-6})}

or q=1.4*10^{-9}C

5 0
3 years ago
A particle released during the fission of uranium-235 is a(n) ________
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<span>A particle released during the fission of uranium-235 is a "Neutron"</span>
6 0
3 years ago
What is the scientific definition of tracked?
dexar [7]
To find or discover by investigation?
8 0
3 years ago
Calculate the electric field at one corner of a square 50 cm on a side if the other corners are occupied by 250x10-7C (charges)
SIZIF [17.4K]

The electric field at one corner of a square is 1614217 N/C.

Explanation:

The distance between x and y direction diagonals.

As per the given details the distance between diagonals is calculated as

0.5² + 0.5² = c²  =>  c = 0.707 m

Charge to the right:  In x direction

In order to find the electric charge towards x direction

we use e = kq/r² formula

As 'k' is coulomb's constant it's value is 9 x 10^{9} N m²/C²

e = (9 x 10^{9})(250 x 10^{-7}) / (0.5)²

e = 9 x 10^{5} N/C

Charge diagonal:

e = kq/r²

e = [(9 x 10^{9})(250 x 10^{-7}) / (0.707)²] cos 45

e = 225000√2 N/C

X direction sum = 1218198 N/C.

Similarly as shown in x direction the charge is same for y direction also

Charge below:  For y direction

e = kq/r²

e = (9 x 10^{9})(250 x 10^{-7}) / (0.5)²

e = 9 x 10^{5} N/C

Charge diagonal:

e = kq/r²

e = [(9 x 10^{9})(250 x 10^{-7}) / (0.5)²] sin 45

e = 159099 N/C

Y direction sum = 1059099 N/C

Resultant electric field strength:

1218198 ² + 1059099² = e²

e = 1614217 N/C [45 degrees below the horizontal]

4 0
3 years ago
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