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Rom4ik [11]
4 years ago
10

carlis lives 100 m away from his friend home what is his average speed if he reaches his friends home in 50s

Physics
1 answer:
Savatey [412]4 years ago
8 0
100m ÷ 50s = 2m/s
Just some simple divison.
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If a 6V battery is connected to a light bulb whose resistance is 55,000Ω How much current will flow in the circuit?
notka56 [123]

Answer:

Current, I = 0.000109 Amps

Explanation:

Given the following data;

Voltage = 6V

Resistance = 55,000 Ohms

To find the current flowing through the circuit;

Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.

Mathematically, Ohm's law is given by the formula;

V = IR

Where;

V represents voltage measured in voltage.

I represents current measured in amperes.

R represents resistance measured in ohms.

Making current the subject of formula, we have;

I = \frac {V}{R}

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I = \frac {6}{55000}

Current, I = 0.000109 Amps

5 0
3 years ago
A student swings a container of water in a vertical circle of radius 1.0 m. Calculate the minimum speed of the container so that
12345 [234]

Answer:

Explanation:

The centripetal acceleration requirement must equal gravity at the top of the circle

mg = mv²/R

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Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of
wariber [46]

Answer:

For left = 0  N/C

For right = 0  N/C

At middle = -7.6836 * 10^{-11} \vec{i}  N/C

Explanation:

Given data :-

б =6.8 * 10^{-22} C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

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E = \frac{\sigma }{2\varepsilon }

1) To the left of the plate

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

2) To the right of them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

3) Between them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(-\vec{i}) = (\frac{\sigma }{\varepsilon })(-\vec{i}) = -\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} }  \vec{i} =   -7.6836 * 10^{-11} \vec{i} N/C

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