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lesya [120]
2 years ago
7

Calculate the reading on voltmeter v²​

Physics
2 answers:
Olegator [25]2 years ago
6 0

Answer:

V

1

and V

2

are equal and out of phase with each other and thus get cancelled. Thus V

3

=220V and i=

100

220

=2.2A

Yuliya22 [10]2 years ago
3 0

The reading of the voltmeter can be determined by finding the potential difference across the 2Ω resistance by using the value of current in the circuit. V=IR, here V is the potential difference across a resistance R through which a current I is flowing.

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The best and most correct answer among the choices provided by your question is the third choice or letter C.

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3 years ago
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Using software, a simulation of a javelin being thrown both on Earth and on the moon is created. In both cases, the velocity and
mario62 [17]

Horizontal distance covered by a projectile is X = Vix *T

where Vix is the initial horizontal component of velocity and T is time taken by the projectile

Vix = ViCos theta

In question they said that initial velocity and angle is same on earth and moon

so Vix would remains same

now let's see about time taken T

time taken to reach the highest point

Vfy = Viy +gt

at highest point vertical velocity become zero so Vfy =0

0 = Vi Sin theta + gt

t = Vi Sintheta /g

Total time taken to land will be twice of that

On earth

Te= 2t

Te = 2Sinθ/g

on moon g is one-sixth of g(earth)

Tm = 2Sinθ/(g/6)

Tm = 6(2Sinθ/g)

Tm = 6Te

so total time taken by the projectile on moon will be six times the time taken on earth

From first equation X = Vix*T

we can see that X will also be 6 times on moon than earth

so projectile will cover 6 times distance on moon than on earth

4 0
3 years ago
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Based on what you learn about jovian moons by watching the videos or reading your textbook, what types of evidence for recent or
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Answer:

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7 0
3 years ago
A man wishes to lift a stone weighing 1440 N, using a first-class lever that measures 5 meters. What force should it perform if
Crazy boy [7]

Answer:

3360 N

Explanation:

In a first-class lever, the effort force and load force are on opposite sides of the fulcrum.

The lever is 5 m long.  The load force is 1.50 m from the fulcrum, so the effort force must be 3.50 m from the fulcrum.

The torques are equal:

Fr = Fr

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4 0
3 years ago
Three charges are located at a different position in a plane: q1= 10μC at →r1=(5,6)cm q2=−27μC at →r2=(−6,10)cm and q3=−12μC at
sasho [114]

Answer:

 E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C

 E_{total} = 2,467 10⁶ N / A       θ = -21.8      

Explanation:

For this exercise we will use that the electric field is a vector quantity, so the total field is

        E_total = E₁₃ + E₂₃

bold font vectors .  We can work with the components of the electric field in each axis

X- axis

       E_ total x = E₁₃ₓ + E_{23x}

y-axis  

      E_{total y} = E_{13y} + E_{23y}

the expression for the electric field is

       E = k q / r²

where r is the distance between the charge and the positive test charge

       

in this exercise

Let's find the field created by charge 1

q₁ = 10 μC = 10 10⁻⁶ C

x₁ = 5 cm = 0.05 m

x₃ = 21 cm = 0.21 m

         E_{13x} = 9 10⁹ 10 10⁻⁶ / (0.21 -0.05)²

         E_{13x} = 3.516 10⁶ N / C

y₁ = 6 cm = 0.06 cm

y₃ = -12 cm = -0.12 m

        E_{13y} = 9 10⁹ 10 10⁻⁶ / (-0.12 - 0.06)²

        E_{13y} = 2,777 10⁶ N / C

let's find the field produced by charge 2

q₂ = -27 μC = - 27 10⁻⁶ C

x₂ = -6 cm = -0.06 m

x₃ = 0.21 m

        E_{23x} = 9 10⁹ 27 10⁻⁶ / (0.21 + 0.06)²

        E_{23x} = 1.23 10⁶ N / A

y₂ = 10 cm = 0.10 m

y₃ = -0.12 m

        E_{23y} = 9 10⁹ 27 10⁻⁶ / (-0.12 - 0.10)²

        E_{23y} = 1.86 10⁶ N / C

Taking the components we can calculate the total electric field, we must use that charge of the same sign repel and attract the opposite sign, remember that the test charge is always considered positive.

       E_{total x} = E_{13x} - E_{23x}

       E_{total x} = (3.516 - 1.23) 10⁶

       E_{total x} = 2.29 10⁶ N / A

       

       E_{total y} = -E_{13y} + E_{23y}

       E_{total y} = (-2.777 +1.86) 10⁶ N / A

       E_{total y} = -0.917 10⁶ N / A

we can give the result in two ways

         E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C

or in the form of modulus and angle, let's use the Pythagorean theorem to find the modulus

                E_{total} = √ (E_{total x}^2 + E_{total y}^ 2)

                 E_{total} = √ (2.29² + 0.917²) 10⁶

                E_{total} = 2,467 10⁶ N / A

let's use trigonometry for the angle

                tan θ = E_total and / E_totalx

                θ = tan⁻¹ E_{total y} / E_{total x}

                θ = tan⁻¹ (-0.917 / 2.29)

                θ = -21.8

The negative sign indicates that the angle is measured with respect to the x-axis in a clockwise direction.

7 0
3 years ago
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