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DedPeter [7]
3 years ago
13

Where is the electric field zero? a. region 2, 0.46 m from the +7 µC charge b. region 2, 0.46 m from the +5 µC charge c. the fie

ld is not zero anywhere on the x axis (except at infinity)d. region 3, 0.46 m from the +7 µC chargee. region 1, 0.46 m from the +5 µC charge
Physics
1 answer:
Ratling [72]3 years ago
3 0

Answer:

the field is not zero anywhere on the x axis (except at infinity)

Explanation:

From the Coulomb's law we have electric field intensity as:

E=\frac{1}{4\pi.\epsilon_o} \frac{q}{r^2}

where:

\epsilon_o= permittivity of free space

q= charge due to which the field is generated.

r= distance from the charge

So, from the above relation:

Electric field due to a charge can only be zero at infinity.

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How do you solve this???
Tom [10]
Rt= ΣR = 40Ω
Vt= 80V
It= 80V/40Ω= 2A
V1= 15Ω*2A= 30V
V2= 20Ω*2A= 40V
V3= 5Ω*2A= 10V
4 0
3 years ago
A man lifts a 25.9 kg bucket from a well and does 5.92 kJ of work.The acceleration of gravity is 9.8 m/s^2 . How deep is the wel
Jet001 [13]

Answer:

The well is 23.3 m

Explanation:

As the bucket is lifted out of the well, energy in the man is being transferred to the bucket as gravitational potential energy.

Work done against gravity =  mass * height * acceleration due to to gravity

W  =  mgh

5 920 J = 25.9 kg * h * 9.8 m/s²

h = 23.3 m

4 0
3 years ago
What is the magnitude of the sum of the two vectors A = 36 units at 53 degrees, and B =47 units at 157 degrees.
Thepotemich [5.8K]

Answer:

51.82

Explanation:

First of all, let's convert both vectors to cartesian coordinates:

Va = 36 < 53° = (36*cos(53), 36*sin(53))

Va = (21.67, 28.75)

Vb = 47 < 157° = (47*cos(157), 47*sin(157))

Vb = (-43.26, 18.36)

The sum of both vectors will be:

Va+Vb = (-21.59, 47.11)   Now we will calculate the module of this vector:

|Va+Vb| = \sqrt{(-21.59)^2+(47.11)^2}=51.82

4 0
4 years ago
WILL GIVE BRAINLIEST!!<br> What is a way to transfer charge in which an object becomes polarized?
Andreyy89

Answer:

I answered Number 4 (Solids and Elasticity)

Explanation:

solids and elasticity

4 0
3 years ago
A car is traveling north at 17.7 m/s . After 6 it’s velocity is 141 in the same direction. Find the magnitude and direction of t
Furkat [3]

By equation of motion we have   v = u + at

Where u = Initial velocity, v = final velocity, t = time taken and a = acceleration

Here v = 141 m/s, u = 17.7 m/s and t = 6 s

On substitution we will get

        141 = 17.7+ 6a

       So, a = (141-17.7)/6 = 20. 55 m/s^{2}

       Aceeleration = 20. 55 m/s^{2} along north direction.


3 0
3 years ago
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