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xxTIMURxx [149]
2 years ago
9

What would be the magnitude of the electric field 0.75 m from a 0.63 C master charge and what would be the force on a 0.50 C tes

t charge placed here?
Physics
1 answer:
bagirrra123 [75]2 years ago
8 0

The magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.

<h3>Electric field on the master charge</h3>

E = kq/r²

where;

  • q is magnitude of master charge
  • r is distance of separation
  • k is Coulomb's constant

E = (9 x 10⁹ x 0.63)/(0.75²)

E = 1.008 x 10¹⁰ N/C

<h3>Force on the test charge</h3>

F = Eq

where;

  • E is electric field
  • q is the test charge

F = (1.008 x 10¹⁰) x (0.5)

F = 5.04 x 10⁹ N

Thus, the magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

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sergiy2304 [10]

Answer:

A kilowatt (kW) is a unit of power.

Explanation:

The power of an object is given by :

P=\dfrac{E}{t}

Here,

E is the energy required

t is time

The SI unit of power is Watts and the SI unit of energy is Joule. the commercial unit of energy is kilowatt per hour.    

Option (1) :  A kilojoule (kJ) is a unit of power is incorrect.

Option (2) : A gigawatt (GW) is a unit of energy is incorrect.

Option (3) : A watt (W) is a unit of energy is incorrect.

Option (4) : A kilowatt x hour per year (kWh/yr) is a unit of energy is incorrect.

Option (4) : A kilowatt (kW) is a unit of power is correct.

Hence, the correct option is (d).

3 0
3 years ago
Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude
labwork [276]

Answer:\tau=1.03\times 10^{-4}\ N-m

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

\tau=NIAB\ \sin\theta

Let us assume that, A=0.0008\ m^2

\theta is the angle between normal and the magnetic field, \theta=90^{\circ}-30^{\circ}=60^{\circ}

Torque is given by :

\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m

So, the net torque about the vertical axis is 1.03\times 10^{-4}\ N-m. Hence, this is the required solution.

4 0
3 years ago
Which of the following is NOT true? a. The higher the moment of inertia, the greater the resistance to changes in angular veloci
kaheart [24]

Answer:

Explanation:

The moment of inertia is the integral of the product of the squared distance by the mass differential. Is the mass equivalent in the rotational motion

a) True. When the moment of inertia is increased, more force is needed to reach acceleration, so it is more difficult to change the angular velocity that depends proportionally on the acceleration

b) True. The moment of inertia is part of the kinetic energy, which is composed of a linear and an angular part. Therefore, when applying the energy conservation theorem, the potential energy is transformed into kinetic energy, the rotational part increases with the moment of inertia, so there is less energy left for the linear part and consequently it falls slower

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Sally Sizzle builds a circuit by connecting three resistors in parallel with a 9V battery. The current in this circuit is 3.6 am
Andru [333]

Then answer is C 9 ohms

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3 years ago
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26. A baseball traveling 38 m/s is caught by the catcher. The catcher takes 0.1 seconds to stop the ball. What is the accelerati
Furkat [3]

Answer:

The ball has an acceleration of -380 m/s², this means the ball slows down

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Explanation:

Step 1: Data given

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At time t, the ball stops. This means v = 0

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Step 2: Calculate the acceleration

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with v0 = the velocity of the ball at time t=0. v0 = 38 m/s

with a= the acceleration in m/s²

with t = time in seconds

0 = 38 + a*0.1

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An acceleration of -380 m/s² is the equivalent of 38.736 g's

5 0
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