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gayaneshka [121]
3 years ago
5

A current of 0.2A flows through a component with a resistance of 40ohms calculate the potential difference

Physics
1 answer:
erastova [34]3 years ago
4 0

Answer:

8V

Explanation:

Given parameters:

Current = 0.2A

Resistance  = 40ohms

Unknown:

Potential difference = ?

Solution:

Potential difference is force that drives current in an electric circuit.

According to Ohms law, it is given as;

   V = IR

V is the potential difference

I is the current

R is the resistance

 Now, insert the parameters and solve;

    V = 0.2 x 40  = 8V

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Why is it important to use the correct number of significant digits when
Artemon [7]

Answer:

D

Explanation:

Scientists use significant figures to avoid claiming more accuracy in a calculation than they actually know.

6 0
3 years ago
Read 2 more answers
A janitor standing on the top floor of a building wishes to determine the depth of the elevator shaft. They drop a rock from res
Bumek [7]

Answer:

Part a)

H = 26.8 m

Part b)

error = 7.18 %

Explanation:

Part a)

As the stone is dropped from height H then time taken by it to hit the floor is given as

t_1 = \sqrt{\frac{2H}{g}}

now the sound will come back to the observer in the time

t_2 = \frac{H}{v}

so we will have

t_1 + t_2 = 2.42

\sqrt{\frac{2H}{g}} + \frac{H}{v} = 2.42

so we have

\sqrt{\frac{2H}{9.81}} + \frac{H}{336} = 2.42

solve above equation for H

H = 26.8 m

Part b)

If sound reflection part is ignored then in that case

H = \frac{1}{2}gt^2

H = \frac{1}{2}(9.81)(2.42^2)

H = 28.7 m

so here percentage error in height calculation is given as

percentage = \frac{28.7 - 26.8}{26.8} \times 100

percentage = 7.18

5 0
3 years ago
If the voltage in a wire is 10 volts, and the current is 5 amps, what is the resistance in the wire? Show work. Please put units
aniked [119]

Answer:2 ohm

Explanation:

5 0
4 years ago
Zona onde não incide luz visível devido à interposição de um objeto opaco.
m_a_m_a [10]

Answer:

can you please translate to english?

Explanation:

8 0
3 years ago
14. Three identical light bulbs are connected in series, then are disconnected and arranged in parallel. For each of the scenari
Inessa [10]
In order to make things easier to describe and explain, let's call
the resistance of each bulb 'R', and the battery voltage 'V'.

a).  In series, the total resistance is 3R.
In parallel, the total resistance is R/3.
Changing from series to parallel, the total resistance of the circuit
decreases to 1/9 of its original value.

b).  In series, the total current is  V / (3R) .
In parallel, the total current is  3V / R .
Changing from series to parallel, the total current in the circuit
increases to 9 times its original value.

c).  In series, the power dissipated by the circuit is 

                                   (V) · V/3R  =  V² / 3R .

In parallel, the power dissipated by the circuit is

                                   (V) · 3V/R  =  3V² / R .

Changing from series to parallel, the power dissipated by
the circuit (also the power delivered by the battery) increases
to 9 times its original value.

8 0
3 years ago
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