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dexar [7]
2 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]2 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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The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine th
Luba_88 [7]

Here is the complete question

The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine the horizontal shear stress at point H, which is located L  = 20 mm below the centriod

The missing image which is the remaining part of this question is attached in the image below.

Answer:

The horizontal shear stress at point H is  \mathbf{\tau_H \approx  42.604 \ N/mm^2}

Explanation:

Given that :

The internal shear force V  =  80 kN = 80 × 10³ N

The moment of inertia = 64,900,000

The length = 20 mm below the centriod

The horizontal shear stress  \tau can be calculated by using the equation:

\tau = \dfrac{VQ}{Ib}

where;

Q = moment of area above or below the point H

b = thickness of the beam = 10  mm

From the centroid ;

Q = Q_1 + Q_{2}

Q = A_1y_1 + A_{2}y_{2}  

Q = ( ( 70 × 10) × (55) + ( 210 × 15) (90 + 15/2) ) mm³

Q = ( ( 700) × (55) + ( 3150 ) ( 97.5)  ) mm³

Q = ( 38500 +  307125 ) mm³

Q = 345625 mm³

\tau_H = \dfrac{VQ}{Ib}

\tau_H = \dfrac{80*10^3  * 345625}{64900000*10 }

\tau_H = \dfrac{2.765*10^{10}}{649000000 }

\tau_H = 42.60400616 \ N/mm^2

\mathbf{\tau_H \approx  42.604 \ N/mm^2}

The horizontal shear stress at point H is  \mathbf{\tau_H \approx  42.604 \ N/mm^2}

7 0
2 years ago
What is Yusef most likely finding?
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Explanation:

yusef adds all of the values in his data set and then divide by the number of values in the set. the actual density of iron is 7.874 g/ml .

4 0
2 years ago
An electron is moving in the vicinity of a long, straight wire that lies along the z-axis. The wire has a constant current of 8.
viktelen [127]

Answer:

The  force that the wire exerts on the electron is -4.128\times10^{-20}i-6.88\times10^{-20}j+0k

Explanation:

Given that,

Current = 8.60 A

Velocity of electron v= (5.00\times10^{4})i-(3.00\times10^{4})j\ m/s

Position of electron = (0,0.200,0)

We need to calculate the magnetic field

Using formula of magnetic field

B=\dfrac{\mu I}{2\pi d}(-k)

Put the value into the formula

B=\dfrac{4\pi\times10^{-7}\times8.60}{2\pi\times0.200}

B=0.0000086\ T

B=-8.6\times10^{-6}k\ T

We need to calculate the force that the wire exerts on the electron

Using formula of force

F=q(\vec{v}\times\vec{B}

F=1.6\times10^{-6}((5.00\times10^{4})i-(3.00\times10^{4})j\times(-8.6\times10^{-6}) )

F=(1.6\times10^{-19}\times3.00\times10^{4}\times(-8.6\times10^{-6}))i+(1.6\times10^{-19}\times5.00\times10^{4}\times(-8.6\times10^{-6}))j+0k

F=-4.128\times10^{-20}i-6.88\times10^{-20}j+0k

Hence, The  force that the wire exerts on the electron is -4.128\times10^{-20}i-6.88\times10^{-20}j+0k

5 0
2 years ago
A goat enclosure is in the shape of a right triangle. One leg of the enclosure is built against the side of the barn. The other
san4es73 [151]

Answer:

16,18,22

Or

1,3,7

Explanation:

The detailed explanation is contained in the image attached. The lengths are found using Pythagoras theorem and the two lengths reflects the two values of x yielded by the quadratic equation

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The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is c
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Answer:

64.59kpa

Explanation:

See attachment

6 0
2 years ago
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