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Gnoma [55]
1 year ago
14

A proton is observed to have an instantaneous acceleration 11*10^11. what is the magnitude of e of the electric field at the pro

ton's location?
Physics
1 answer:
jek_recluse [69]1 year ago
6 0

The magnitude of the electric field at the proton's location is 10,437.5 N/C.

<h3>What the magnitude of the electric field?</h3>

The size of the electric field is basically characterized as the power per charge on the test charge. On the off chance that the electric field strength is meant by the image E. Very much like gravity, electric fields work the same way. In any case, while gravity generally draws in, an electric field, then again, can either rebuff or draw in. By and large, the Electric Field submits to the super-position guideline. the all out Electric Field from various charges is equivalent to the amount of the electric fields from each charge separately. An electric field is the actual field that encompasses electrically charged particles and applies force on any remaining charged particles in the field, either drawing in or repulsing them.

Learn more about the magnitude of the electric field, visit

brainly.com/question/26898699

#SPJ4

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The distance from the Earth to the Sun is 92 868 000 miles.
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Determine
djyliett [7]

(i) The total capacitance for the circuit is 5 μF.

(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.

(iii) The charge stored in 3μF capacitor is  6 x 10⁻⁶ C.

<h3>Total capacitance of the circuit</h3>

The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.

<h3>C1 and C2 are in series </h3>

\frac{1}{C_{12}} = \frac{1}{C_1 } + \frac{1}{C_2} \\\\\frac{1}{C_{12}} = \frac{1}{4 } + \frac{1}{4} \\\\\frac{1}{C_{12}} = \frac{1}{2} \\\\C_{12} = 2 \ \mu F

<h3>C1 and C2 are parallel to C3</h3>

C_{123} = C_{12} + C_3\\\\C_{123} = 2\ \mu F + 2\ \mu F \\\\C_{123} = 4 \ \mu F

<h3>C(123) is series to C5 and C6</h3>

\frac{1}{C_{t} } = \frac{1}{C_{123}} + \frac{1}{C_5} + \frac{1}{C_6} \\\\\frac{1}{C_{t} } = \frac{1}{4} + \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{t} } = \frac{12}{24} \\\\\frac{1}{C_{t} } = \frac{1}{2} \\\\C_t = 2 \ \mu F

<h3>C7 and C8 are in series</h3>

\frac{1}{C_{78}} = \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{78}} = \frac{2}{6} \\\\\frac{1}{C_{78}} =\frac{1}{3} \\\\C_{78} = 3 \ \mu F

<h3>Total capaciatnce of the circuit</h3>

Ct + C(78) = 2 μF + 3 μF = 5 μF

<h3 /><h3>Total charge stored in the circuit</h3>

The total charge stored in the capacitor is calculated as follows;

Q = CV

Q = (5 x 10⁻⁶) x (20)

Q = 1 x 10⁻⁴ C

<h3>Charge stored in 3μF capacitor</h3>

Q =  (3 x 10⁻⁶) x (20)

Q = 6 x 10⁻⁶ C

Learn more about capacitance of capacitor here: brainly.com/question/13578522

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What occurs during stage one sleep
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Answer:

Stage one of sleep, also known as the transitional phase, occurs when one finds themselves floating in and out of consciousness. During this NREM stage, you may be partially awake while your mind begins to drift off. This period of drowsiness eventually leads to a light sleep

Explanation:

i found it on google

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