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dexar [7]
3 years ago
15

A 120-V electrical line in a home is connected to a 100-W lightbulb, a 180-W television set, a 230-W desktop computer, a 1050-W

toaster, and a 240-W refrigerator.
How much current is flowing in the line?

Express your answer with the appropriate units.
Physics
1 answer:
Natasha_Volkova [10]3 years ago
5 0

Answer:

Therefore,

15 Ampere current is flowing in the line.

Explanation:

Given:

Voltage, V = 120 V

Power of Appliances,

P1 = 100-W lightbulb,

P2 = 180-W television set,

P3 = 230-W desktop computer,

P4 =  1050-W toaster, and

P5 = 240-W refrigerator.

To Find:

Current  flowing in the line, I = ?

Solution:

We have Power formula,

Power=Voltage\times Current\\P=V\times I

For Household Voltage Remain Same and Current will differ and there will be a Parallel Connection for the Appliances,

Therefore,

1 .For lightbulb,

P_{1}=V\times I_{1}

Substituting the values we get

100 = 120\times I_{1}\\I_{1}=0.8333\ Ampere

2. Similarly for television,

P_{2}=V\times I_{2}

Substituting the values we get

180= 120\times I_{2}\\I_{2}=1.5\ Ampere

3. Similarly for desktop computer,

P_{3}=V\times I_{3}

Substituting the values we get

230= 120\times I_{3}\\I_{3}=1.9166\ Ampere

4. Similarly for toaster,

P_{4}=V\times I_{4}

Substituting the values we get

1050= 120\times I_{4}\\I_{4}=8.75\ Ampere

5. Similarly for refrigerator,

P_{5}=V\times I_{5}

Substituting the values we get

240= 120\times I_{5}\\I_{5}=2\ Ampere

In Parallel connection Total current 'I' is given as,

I=I_{1}+I_{2}+I_{3}+I_{4}+I_{5}

Substituting the values we get

I=0.8333+1.5+1.9166+8.75+2=14.9999\approx 15\ Ampere

Therefore,

15 Ampere current is flowing in the line.

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A 20 cm-long wire carrying a current of 6 A is immersed in a uniform magnetic field of 3 T. If the magnetic field is oriented at
SOVA2 [1]

Answer:

the  magnitude of the force that the wire will experience = 1.8 N

Explanation:

The force on a current carrying wire placed in a magnetic field is :

F = Idl × B

where:

I = current flowing through the wire

dl = length of the wire

B = magnetic field

We can equally say that :

|F| = IdlBsin \theta

where : sin θ is the angle at which the orientation from the magnetic field  to the wire occurs = 30°

Then;

|F| = B\ I \ L \ sin \theta

Given that:

L = 20 cm = 0.2 m

I = 6 A

B = 3 T

θ = 30°

Then:

F = 3 × 6 × 0.2 sin 30°

F = 1.8 N

Therefore, the  magnitude of the force that the wire will experience = 1.8 N

6 0
3 years ago
True or False?
WINSTONCH [101]

Answer:

true is the answer of the question

5 0
3 years ago
Read 2 more answers
¿cual es la velocidad de un haz de electrones que marchan sin desviarse cuando pasan a traves de un campo magnetico perpendicula
Elina [12.6K]

Answer:

La velocidad del haz de electrones es 1.78x10⁵ m/s. Este valor se obtuvo asumiendo que el campo magnético dado (3500007) estaba en tesla y que la fuerza venía dada en nN.

Explanation:

Podemos encontrar la velocidad del haz de electrones usando la Ley de Lorentz:

F = |q|vBsin(\theta)     (1)

En donde:

F: es la fuerza magnética = 100 nN

q: es el módulo de la carga del electron = 1.6x10⁻¹⁹ C

v: es la velocidad del haz de electrones =?

B: es el campo magnético = 3500007 T

θ: es el ángulo entre el vector velocidad y el campo magnético = 90°

Introduciendo los valores en la ecuación (1) y resolviendo para "v" tenemos:

v = \frac{F}{qBsin(\theta)} = \frac{100 \cdot 10^{-9} N}{1.6 \cdot 10^{-19} C*3500007 T*sin(90)} = 1.78 \cdot 10^{5} m/s            

Este valor se calculó asumiendo que el campo magnético está dado en tesla (no tiene unidades en el enunciado). De igual manera se asumió que la fuerza indicada viene dada en nN.

Entonces, la velocidad del haz de electrones es 1.78x10⁵ m/s.  

Espero que te sea de utilidad!                                        

7 0
3 years ago
The driver of a car slams on the brakes when he sees a tree blocking the road. the car slows uniformly with acceleration of -5.9
Hatshy [7]
Let u =  the speed of the car at the instant when braking begins.

The braking distance is s = 62.3 m, the acceleration is a = -5.9 m/s², and the braking duration is t = 4.15 s.

Use the formula s = ut + (1/2)at² to obtain
(u m/s)*(4.15 s) + 0.5*(-5.9 m/s²)*(4.5 s)² = (62.3 m)
4.15u = 62.3 + 50.8064 = 113.1064
      u = 27.2546 m/s

Let v m/s be the speed with which the car strikes the tree.
Then
v = 27.2546 - 5.9*4.15
   = 2.7696 m/s

Answer: 2.77 m/s (nearest hundredth)

4 0
3 years ago
A large mass has (less or more) mass energy than a smaller mass at the same temperature?
Alexeev081 [22]

Answer:

more I guess

Explanation:

hope this help

4 0
3 years ago
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