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Hoochie [10]
3 years ago
10

Which factor indicates the amount of charge on the source charge?

Physics
1 answer:
Westkost [7]3 years ago
3 0

Answer:

B. the number of field lines on the source charge

Explanation:

As we know that electric flux is defined as the number of electric field lines passing through a given area.

So here  electric flux due to a point charge "q" is given by

so here we know that flux depends on the magnitude of charge and hence we can say that number of filed lines originating from a point charge will depends on the magnitude of the charge.

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Given: Q1 = +10 uc = 1.0 x 10-5C
ser-zykov [4K]

Answer:

-0.038 N

Explanation:

F=K Q1 Q2/r^2 by COULOMB'S LAW

F= 9×10^9×1×10^-5×-1.5×10^-5/(6)^2

F= -0.038 N

5 0
3 years ago
How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t
Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

Qo=1.6 \times 10^{2} \mathrm{~mL}

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}

Generally, the equation for Rate of flow of Liquid is  mathematically given as

\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}

$$

Where dP is pressure difference r is the radius

\mu is the viscosity of water

L is the length of the pipe

Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}

Q=5.2 \mathrm{~mL} / \mathrm{s}

In $30s the quantity that flows out of the tube

&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}

In conclusion, the quantity that flows out of the tube

Qo=1.6 \times 10^{2} \mathrm{~mL}

Read more about the flows rate

brainly.com/question/27880305

#SPJ1

5 0
2 years ago
Two circular disks spaced 0.50 mm apart form a parallel-plate capacitor. transferring 1.50 ×109 electrons from one disk to the o
blondinia [14]

The diameter of the disks is 1.32 cm.

If n electrons each of charge e are transferred from one disc to another, calculate the total charge Q transferred from one disc to the other using the expression,

Q=ne

Substitute 1.50×10⁹ for n and 1.6×10⁻¹⁹C for e.

Q=ne\\ =(1.50*10^9)(1.6*10^-^1^9C)\\ =2.4*10^-^1^0C

The potential difference V between the disks separated by a distance d is given by,

V=Ed

here, E is the electric field.

Substitute 2.00×10⁵N/C for E and 0.50×10⁻³m for d.

V=Ed\\ =(2.00*10^5N/C)(0.50*10^-^3m)\\ =100V

The capacitance C of the capacitor is given by,

C=\frac{Q}{V} \\ =\frac{(2.4*10^-^1^0C)}{100V} \\ =2.4*10^-^1^2F

The capacitance of a parallel plate capacitor is given by,

C=\frac{\epsilon_0A}{d}

Here, ε₀ is the permittivity of free space  and A is the area of the disks.

Rewrite the expression for A.

C=\frac{\epsilon_0A}{d}\\ A=\frac{Cd}{\epsilon_0} \\ =\frac{(2.4*10^-^1^2F)(0.50*10^-^3m)}{(8.85*10^-^1^2C^2/Nm^2)} \\ =1.36*10^-^4m^3

the area A of the disks is given by,

A=\frac{\pi D^2}{4} \\ D=\sqrt{\frac{4A}{\pi } }

Here, D is the diameter of the disk.

D=\sqrt{\frac{4A}{\pi } }\\ =\sqrt{\frac{4(1.36*10^-^4m^2)}{3.14} } \\ =0.01316m\\ =1.32cm

The diameter of each disc is found to be 1.32 cm.



7 0
3 years ago
PLEASE HELP! 2ND TIME I ASKED THIS!!!
Goshia [24]

The train is moving at 50 m/s and Emma is walking down the aisle with 1 m/s speed in the same direction of train. The relative velocity of Emma with respect to other passengers pf the train would be 1 m/s. This is because, the train is not moving relative to them and only emma is moving at 1 m/s. If a person observes from outside, Emma would have (50 +1) m/s = 51 m/s velocity.

relative velocity when two objects are moving in same direction as oberved from outside observer:

v_{ab}=v_a+v_b

relative velocity when two objects are moving in the opposite directionas oberved from outside observer:

v_{ab}=v_a-v_b

4 0
3 years ago
A 0.54 kg particle has a speed of 5.0 m/s at point A and kinetic energy of 7.5 J at point B. What is
Wewaii [24]

<u>Answers</u>

(a)  6.75 Joules.

(b)  5.27 m/s

(c) 0.75 Joules


<u>Explanation</u>

Kinetic energy is the energy possessed by a body in motion.

(a) its kinetic energy at A?

K.E = 1/2 mv²

       = 1/2 ×  0.54 × 5²

       = 6.75 Joules.

(b) its speed at point B?

K.E = 1/2 mv²

7.5 = 1/2 × 0.54 × V²

V² = 7.5 ÷ 0.27

     = 27.77778

V = √27.77778

   = 5.27 m/s

(c) the total work done on the particle as it moves from A to B?

Work done = 7.5 - 6.75

                 = 0.75 Joules

4 0
3 years ago
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