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iris [78.8K]
3 years ago
12

Answer True or Flase1-Electric potential due to a uniform E field doesn’t change with location.2-The equipotential surfaces asso

ciated with a uniform E field consist of a family of planes that are perpendicular to the E field.3-The electric potential difference between two locations can exist when there is only a source charge. Even when there is no TEST charge q, the potential difference can exist between two points in the space.
Physics
1 answer:
TEA [102]3 years ago
3 0

Answer:

1. False

2. True

3. True

Explanation:

1- False —> The relation between electric potential and electric field is given such that

-\int\limits^a_b \vec{E}d\vec{l} = V_{ab}

Therefore, for a uniform E field, electric potential is linearly proportional to the distance.

2- True —> The electric field lines always cross the equipotential lines perpendicularly.

3- True —> In order to be a potential difference, one source of electric field is enough. The electric potential will decrease radially according to the following formula:

V = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}

There is no test charge in the formula, only the source charge. Even when there is no test charge, the potential difference between points in space can exist.

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Water flows through a horizontal pipe with a cross-sectional area of 3mat a speed of 10 m/s with a pressure of 3 Bars at a Point
Over [174]

(a) The speed of the water at point B  is 10 m/s and the speed at point C is 5 m/s.

(b) The pressure of the water at point B is 4.96 bars and the pressure at point C is 5.335 bars.

<h3>Speed of the water at point B </h3>

The speed of the water at point B is determined from continuity equation as shown below;

Q = AV

A₁V₁ = A₂V₂

Since, A₁ = A₂ =  3 m²

V₁ = V₂ = 10 m/s

Thus, the speed of the water at point B  is 10 m/s.

<h3>Speed of water at point C</h3>

A₁V₁ = A₃V₃

A₁V₁  = (2A₁)V₃

V₃ = (A₁V₁) / 2A₁

V₃ = V₁ / 2

V₃ = 10/2

V₃ = 5 m/s

Thus, the speed of the water at point C  is 5 m/s.

<h3>Pressure of the water at point B </h3>

The pressure of the water at point B is determined by appying Bernoulli's equation as shown below;

P₁ + ¹/₂ρV₁² + ρgh₁ = P₂ + ¹/₂ρV₂² + ρgh₂

where;

  • P₁ = 3 bars = 300,000 Pa
  • ρ density of water = 1000 kg/m³
  • g is acceleration due to gravity = 9.8 m/s²

300,000 + ¹/₂(1000)(10)² + (1000)(9.8)(20) = P₂ + ¹/₂(1000)(10)²

546,000 = P₂ + 50,000

P₂ = 496,000 Pa

P₂ = 4.96 bars

<h3>Pressure of the water at point C</h3>

P₁ + ¹/₂ρV₁² + ρgh₁ = P₃ + ¹/₂ρV₃² + ρgh₃

546,000 = P₃ + ¹/₂(1000)(5)²

546,000 = P₃ + 12,500

P₃ = 533,500 Pa

P₃ = 5.335 bars

Learn more about pressure in horizontal pipe here: brainly.com/question/26761275

8 0
2 years ago
A wire with radius 23 cm has a current of 7 A which is distributed uniformly through its cross sectional area. If you were to us
Rina8888 [55]

Answer:

The magnetic induction of the magnetic field is  0.0005293 mT

Explanation:

Data given

I = 7 A = the total current in the wire

r = 23 cm = the radius of the wire = 0.23 meter

r' = 2cm = the measurement point, which should be inside the wire = 0.02 meter

Let's consider the current density is constant in the wire, ⇒  the current enclosed is a function of the enclosed area

I(enclosed) = Jπ r ²

we can  consider the current density  as the total current over the whole area:

I(enclosed) = I / (πr ²)  * πr' ²

I(enclosed) = (I* r'²)/ (r ²)  

with I =  total current in the wire = 7A

With r = the radius of wire = 0.23 meter

with r' = the distance of point from the center of wire  0.02 meter

We plug this into ampere's law:

∮ *B *dl =μ 0  * (I* r'²)/ (r ²)  

with B = Magnetic flux density (in Tesla) or magnetic induction

with dl = an infinitesimal element (a differential) of the curve C

with µ0 = the magnectic constant =  4π*10^−7 H/m

We can simplify this, by using an Amperian loop can write this as:

B *( 2 π r') =  μ 0  * (I* r'²)/ (r ²)  

Because the circumference of a circle is  2 π r , when we integrate over length at a distance  r ′  from the center of wire whose crossection is a circle we get  2 π r ′

When we isolate B, we get:

B = µo *(Ir'/2 π r ²)

B =  4π*10^−7 * ((7*0.02)/2*π*0.23²)

B =5.293 *10 ^-7 T  = 0.0005293 mT

The magnetic induction of the magnetic field is  0.0005293 mT

6 0
3 years ago
A marble slides without friction in a vertical loop around the inside of a smooth, 28.6 cm diameter horizontal pipe. The marble'
Leviafan [203]

Answer:3.49 m/s

Explanation:

Given

Speed of marble at Bottom v=4.22 m/s

Diameter of loop d=28.6 cm

As Energy is conserved therefore Energy at top is equal to energy at bottom

E_T=E_B

\frac{mv^2}{2}+mgh=\frac{mv_0^2}{2}  ,where v_0 is the velocity at bottom

\frac{v^2}{2}+gh=\frac{v_0^2}{2}

v_0^2=v^2+2gh

v^2=v_0^2-2gh

v=\sqrt{v_0^2-2gh}

v=\sqrt{4.22^2-2\times 9.8\times 0.286}

v=\sqrt{17.8084-5.6056}

v=3.49 m/s

                       

7 0
3 years ago
How does 625 over 1000 equal 5/8
ki77a [65]
625 /5=125/5=25/5=5
1000/5=200/5=40/5=8
is 5 Over 8
5 0
3 years ago
Hey! Can someone help with this question? Thx :)
Schach [20]
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