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kozerog [31]
2 years ago
10

An electret is similar to a magnet, but rather than being permanently magnetized, it has a permanent electric dipole moment. Sup

pose a small electret with electric dipole moment 1.0 × 10−7 Cm is 25 cm from a small ball charged to + 80 nC, with the ball on the axis of the electric dipole.What is the magnitude of the electric force on the ball?
Physics
2 answers:
steposvetlana [31]2 years ago
7 0

Answer:

 F = 9.216 × 10⁻³ N

Explanation:

given,

dipole moment = 1 × 10⁻⁷ Cm

distance apart from +80 nC charge = 25 cm

to calculate the magnitude of electric force

Electric field due to dipole

E = \dfrac{2p}{4\pi \varepsilon_0 r^3 }

E = \dfrac{9\times 10^9\times 2\times 1 \times 10^{-7}}{25^3\times 10^{-6} }

E = 1.152 × 10⁵ N/C

electric force on the ball

F = E q

  = 1.152 × 10⁵ × 80 × 10⁻⁹

 F = 9.216 × 10⁻³ N  

Hence, the electric force is equal to  F = 9.216 × 10⁻³ N

Eduardwww [97]2 years ago
5 0

Answer:

Explanation:

Given that,

Chase Q = 80nC = 80×10^-9C

dipole moment p = 1×10^-7 Cm

Distance z = 25cm = 0.25

The Electric field along a dipole moment is given as

E = p / (2π•εo•z³)

We know that k = 1/4π•εo

Then, Electric field becomes

E = 2kp/r³

Where k = 9×10^9 Nm²/C²

Then, the force is given as

F = qE

Therefore,

F = 2k•q•p/r³

F = 2× 9×10^9 ×80×10^-9×1×10^-7/0.25³

F = 9.216 × 10^-3 N

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How much gravitational potential energy does a 45.2 kg object have when it is 21.9m above the ground?
Blizzard [7]

Answer:

Explanation:

The formula for gravitational potential energy is

Ep = m · g · h   Assuming that the acceleration is g = 10m/s²

Ep = 45.4 · 10 · 21.9 = 9,942.6 J

God is with you!!!

6 0
1 year ago
Betelgeuse is the bright red star representing the left shoulder of the constellation Orion. All the following statements about
Sav [38]

Answer:

Option A; ITS SURFACE IS COOLER THAN THE SURFACE OF THE SUN.

Explanation:

A red supergiant star is a larger and brighter type of red giant star. Red supergiants are often variable stars and are between 200 to 2,000 times bigger than the Sun. Example is Betelgeuse.

Betelgeuse is one of the largest known stars, it has a diameter of about 700 times the size of the Sun or 600 million miles, it emits almost 7,500 times as much energy as the Sun, it has a rather low surface temperature (6000F compared to the Sun's 10,000F); this means that it has a more cooler surface than the Sun's surface.

This low temperature also means that the star will appear orange-red in color, and the combination of size and temperature makes it a kind of star called a red super giant.

Although, all the statements above are correct, the only one that can be inferred from the red color of Betelgeuse is that ITS SURFACE IS COOLER THAN THE SURFACE OF THE SUN.

3 0
2 years ago
The resistivity of a metal increases slightly with increased temperature. This can be expressed as rho=rho0[1+α(T−T0)], where T0
Readme [11.4K]

Answer:

I = ΔVA[1 - α (T₀ - T)]/Lρ₀

Explanation:

We have the following data:

ΔV = Battery Terminal Voltage

I = Current through wire

L = Length of wire

A = Cross-sectional area of wire

T = Temperature of wire, when connected across battery

T₀ = Reference temperature

ρ = Resistivity of wire at temperature T

ρ₀ = Resistivity of wire at reference temperature

α = Temperature Coefficient of Resistance

From OHM'S LAW we know that;

ΔV = IR

I = ΔV/R

but,  R = ρL/A   (For Wire)

Therefore,

I = ΔV/(ρL/A)

I = ΔVA/ρL

but,   ρ = ρ₀[1 + α (T₀ - T)]

Therefore,

I = ΔVA/Lρ₀[1 + α (T₀ - T)]

I = [ΔVA/Lρ₀] [1 + α (T₀ - T)]⁻¹

using Binomial Theorem:

(1 +x)⁻¹ = 1 - x + x² - x³ + ...

In case of [1 + α (T₀ - T)]⁻¹, x = α (T₀ - T).

Since, α generally has very low value. Thus, its higher powers can easily be neglected.

Therefore, using this Binomial Approximation, we can write:

[1 + α (T₀ - T)]⁻¹ = [1 - α (T₀ - T)]

Thus, the equation becomes:

<u>I = ΔVA[1 - α (T₀ - T)]/Lρ₀ </u>

3 0
2 years ago
A particle with a charge of -1.24 x 10"° C is moving with instantaneous velocity (4.19 X 104 m/s)î + (-3.85 X 104 m/s)j. What is
astra-53 [7]

Answer:

(a) F= 6.68*10¹¹⁴ N (-k)

(b) F =( 6.68*10¹¹⁴ i  + 7.27*10¹¹⁴ j  ) N

Explanation

To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:

F=q* v X B Formula (1 )

q: charge (C)

v: velocity (m/s)

B: magnetic field (T)

vXB : cross product between the velocity vector and the magnetic field vector

Data

q= -1.24 * 10¹¹⁰ C

v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j

B  =(1.40 T)i  

B  =(1.40 T)k

Problem development

a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =

            = - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)

1T= 1 N/ C*m/s

We apply the formula (1)

  F= 1.24 * 10¹¹⁰ C*  5.39* 10⁴ m/s* N/ C*m/s (-k)

   F= 6.68*10¹¹⁴ N (-k)

a)  vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =

             =( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T

1T= 1 N/ C*m/s

We apply the formula (1)

F= 1.24 * 10¹¹⁰ C*  (  5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s

F =( 6.68*10¹¹⁴  i  + 7.27*10¹¹⁴  j  ) N

8 0
1 year ago
A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?
ankoles [38]
Given required solution

M=10kg W=? W=Fd
v=5.0m/s F=mg
t=2.40s =10*10=100N
S=VT
=5m/s*2.4s
=12m
so W=12*100
W=1200J
4 0
1 year ago
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