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3 years ago

A car takes 15 minutes to travel along a road that is 20 km long.

Physics

2 answers:

iogann1982 [59]3 years ago

8 0

Answer:

Explanation:

15. 20

x4. x4

60. 80

80km/h

kupik [55]3 years ago

3 0

Answer:

80km/h

Explanation:

<u>solution</u>

given

d=20km

t=15min/60min=0.25h

s=?

s=d/t

s=20km/0.25h

s=80km/h

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There is a current of 0.83 A through a lightbulb in a 120 V circuit. What is the resistance of this lightbulb?

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Considering the Ohm's law, the resistance of the lightbulb is 144.58 Ω.

<h3>Definition of current</h3>

The flow of electricity through an object, such as a wire, is known as current (I). Its unit of measure is amps (A). So the current is a measure of the speed at which the charge passes a given reference point in a specified direction.

<h3>Definition of voltage</h3>

The driving force (electrical pressure) behind the flow of a current is known as voltage and is measured in volts (V) (voltage can also be referred to as the potential difference or electromotive force). That is, voltage is a measure of the work required to move a charge from one point to another.

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Resistance (R) is the difficulty that a circuit opposes to the flow of a current and it is measured in ohms (Ω).

Ohm's law establishes the relationship between current, voltage, and resistance in an electrical circuit.

This law establishes that the intensity of the current that passes through a circuit is directly proportional to the voltage of the same and inversely proportional to the resistance that it presents.

Mathematically, Ohm's law is expressed as:

$I=\frac{V}{R}$

Where:

I is the current measured in amps (A).
V the voltage measured in volts (V).
R the resistance that is measured in ohms (Ω).

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In this case, you know that the voltage between two points in a circuit is 120 V and there is a current of 0.83 A.

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3 years ago

Read 2 more answers

Help!!!, combination circuits, Physics

Kaylis [27]

Current and voltage on each resistor:

$I_1 = 3.98 A, V_1 = 3.98 V$

$I_2=0.015 A, V_2 = 0.075 V$

$I_3 = 0.4 A, V_3 = 0.4 V$

$I_4 = 0.385 A, V_4 = 0.77 V$

$I_5 = 0.585 A, V_5 = 1.17 V$

$I_6 = 3.01 A, V_6 = 6.02 V$

$I_7 = 0.97 A, V_7 = 4.85 V$

Explanation:

In order to solve the circuit, we first have to find the equivalent resistance of the whole circuit, then the total current, and then we can proceed finding the current and the voltage for each resistor.

We start by calculating the equivalent resistance of resistors 2 and 3, which are in parallel:

$R_{23}=\frac{R_2R_3}{R_2+R_3}=\frac{(5)(1)}{5+1}=0.833\Omega$

This resistor is in series with resistor 4, so:

$R_{234}=R_{23}+R_4=0.833+2.0=2.833\Omega$

This resistor is in parallel with resistor 5, therefore:

$R_{2345}=\frac{R_{234}R_5}{R_{234}+R_5}=\frac{(2.833)(2.0)}{2.833+2.0}=1.172\Omega$

This resistor is in series with resistor 7, so:

$R_{23457}=R_{2345}+R_7=1.172+5.0=6.172\Omega$

This resistor is in parallel with resistor 6, so:

$R_{234567}=\frac{R_{23457}R_6}{R_{23457}+R_6}=\frac{(6.172)(2.0)}{6.172+2.0}=1.510\Omega$

Finally, this combination is in series with resistor 1:

$R_{eq}=R_1+R_{234567}=1.0+1.510=2.510\Omega$

We finally found the equivalent resistance of the circuit. Now we can find the total current in the circuit, which is also the current flowing through resistor 1:

$I_1=\frac{V}{R_{eq}}=\frac{10}{2.510}=3.98 A$

And we can also find the potential difference across resistor 1:

$V_1=I_1 R_1=(3.98)(1.0)=3.98 V$

This means that the voltage across resistor 6 is

$V_6=V-V_1=10-3.98=6.02 V$

And so, the current on resistor 6 is

$I_6=\frac{V_6}{R_6}=\frac{6.02}{2.0}=3.01 A$

The current flowing in the whole part of the circuit containing resistors 2,3,4,5,7, and therefore through resistor 7, is

$I_7=I-I_6=3.98-3.01=0.97 A$

And so the voltage across resistor 7 is

$V_7=I_7 R_7=(0.97)(5.0)=4.85 V$

The voltage across resistor 5 is

$V_5 = V_6 - V_7 = 6.02 - 4.85 =1.17 V$

And so the current is

$I_5 = \frac{V_5}{R_5}=\frac{1.17}{2.0}=0.585 A$

The current through resistor 4 is

$I_4 = I_7 - I_5 = 0.97-0.585 = 0.385 A$

And therefore its voltage is

$V_4=I_4 R_4 = (0.385)(2.0)=0.77 V$

So, the voltage through resistor 3 is

$V_3=V_5-V_4=1.17-0.77=0.4 V$

And the current is

$I_3=\frac{V_3}{R_3}=\frac{0.4}{1.0}=0.4 A$

Finally, the current through resistor 2 is

$I_2=I_4-I_3=0.5-0.385=0.015 A$

And so its voltage is

$V_2=I_2R_2=(0.015)(5.0)=0.075 V$

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3 years ago

PLEASE HELP ASAP!!!!!

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3 years ago