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3 years ago

What is the value of V^1?

Physics

1 answer:

Neporo4naja [7]3 years ago

3 0

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Current Electricity.

Since here, A potential divider is setup with two resisters in series, that is 3 ohm and 6 ohm.

As the voltage is always in directly proportional to the resistance,by ohm law,

V = IR, so the increase in Voltage will lead to increase in the resistance too.

hence,

for 6 ohm the voltage is 4 V, then for the 3 ohm it would be ,

V1 = (4/6)*3

==> V1 = 2 Volts,

==> hence the potential V1 = 2V

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What is keeping all variables the same except for the one being tested an example of?

stepan [7]

Answer:

A fair test.

Explanation:

Hi, a fair test is used to do scientifically valuable experiments, is a controlled investigation to answer a scientific question.

In a fair test two or more things are compared.

It consists in changing only one factor (the one bieng tested) and keeping all the other conditions the same during an experiment.

The factor is called a variable.

8 0

3 years ago

Read 2 more answers

What is the mass of an object that experiences a gravitational force of 510 N near Earth's surface?

sineoko [7]

Answer:The mass of an object is 52 kg.

Explanation:

Gravitational force on the object ,F=510 N

Acceleration due to gravity = g = $9.8 m/s^2$

Mass of the object = m

Force = mass × acceleration

$510 N=m\times 9.8 m/s^2$

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The mass of an object is 52 kg.

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4 years ago

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a. When the electric field between the plates is 75% of the dielectric strength and energy density of the stored energy is 2800

Olegator [25]

Answer: The value of the dielectric constant k = 1.8

Explanation:

If C= ε A/d and

Electrostatic energy W = 1/2CV^2

Substitutes C in the first formula into the energy formula.

W = 1/2 ε A/d × V^2

Let us remember that electric field strength E is the ratio of potential V to the distance d. Where V = Ed

Substitute V = Ed into the energy W.

W = 1/2 × ε A/d ×( Ed )^2

W = 1/2 × ε A/d × E^2 × d^2

d will cancel one of the ds

W = 1/2 × ε Ad × E^2

W/Ad = 1/2 × ε × E^2

W/V = 1/2 × ε E^2

Where Ad = volume V

E = dielectric strength

εo = permittivity of free space = 8.84 x 10^-12 F/m

W/V = 2800 J/m^3

Let first calculate the dielectric strength

2800 = 1/2 × 8.84×10^-12 × E^2

5600 = 8.84×10^-12E^2

E^2 = 5600/8.84×10^-12

E = sqrt( 6.3 × 10^14)

E = 25 × 10^7

75% of E = 18.9 × 10^6Jm

The permittivity of the material will be achieved by using the same formula

2800 = 1/2 × ε E^2

2800 = 0.5 × ε × (18.9×10^6)^2

2800 = ε × 1.78 × 10^14

ε = 2800/1.78×10^14

ε = 1.57 × 10^-11

Dielectric constant k = relative permittivity

Relative permittivity is the ratio of the permittivity of the material to the permittivity of the vacuum in a free space. That is

k = 1.57×10^-11/8.84×10^-12

k = 1.776

k = 1.8 approximately

Therefore, the value of the dielectric constant k is 1.8

3 0

3 years ago

The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is

stealth61 [152]

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

$\rho_m = 846lb/ft^3$

$g = 32.17405ft/s^2$

$h_1 = 1in = \frac{1}{12} ft$

For the air the defined properties would be

$\rho_a = 0.0075lb/ft^3$

$g = 32.17405ft/s^2$

$h_2 = ?$

We have for equilibrium that

$\text{Pressure change in Air}=\text{Pressure change in Mercury}$

$\rho_m g h_1 = \rho_a g h_2$

Replacing,

$(846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)$

Rearranging to find $h_2$

$h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}$

$h = 9400ft$

Therefore the elevation of the mountain top is 9400ft

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3 years ago

DESCRIBE THE FORMATION OF THE SOLAR SYSTEM ACCORDING TO THE NEBULAR THEORY

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3 years ago