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nadezda [96]
3 years ago
5

The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the gr

ound. To maximize the potential difference between one end of the fluorescent tube and the other, how should the tube be held? Select the best answer from the choices provided. A. The tube should be held horizontally, parallel to the ground. B.The potential difference between the ends of the tube does not depend on the tube's orientation. C.The tube should be held vertically, perpendicular to the ground.
Physics
1 answer:
s344n2d4d5 [400]3 years ago
7 0

Answer:C.The tube should be held vertically, perpendicular to the ground.

Explanation: Power lines are mainly overhead power installation that transfer electric energy from one place to another. Electric power lines contains both Magnetic field and Electric field.

Potential difference is the change in the amount of energy carried by an electric circuit from one point to another. TO MAXIMIZE THE POTENTIAL DIFFERENCE BETWEEN ONE END OF THE TUBE AND ANOTHER? THE TUBE SHOULD BE HELD VERTICALLY,PERPENDICULAR TO THE GROUND.

You might be interested in
Where coastal rock is softer, waves erode the land faster ​
prohojiy [21]

Answer:

Headlands and bays are created where there are bands of hard and soft rock which meet the coastline at right angles. Softer rock is eroded more quickly and erodes backwards to form bays (which may have beaches). The harder rocks are more resistant to erosion and jut out into the sea to form exposed headlands

5 0
2 years ago
Three identical resistors are connected in parallel to a battery. If the current of 12. A flows from the battery, how much curre
Doss [256]

Answer:

4 A

Explanation:

We are given that

R_1=R_2=R_3=4\Omega

I=12 A

We have to find the current flowing through each resistor.

We know that in parallel combination current flowing through different resistors are different and potential difference across each resistor is same.

Formula :

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}

Using the formula

\frac{1}{R}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}

R=\frac{4}{3}\Omega

V=IR

Substitute the values

V=12\times \frac{4}{3}=16 V

I_1=\frac{V}{R_1}=\frac{16}{4}=4 A

I_1=I_2=I_3=4 A

Hence, current flows through any one of the resistors is 4 A.

7 0
2 years ago
An electron enters the gap between the plates of a capacitor at the center of the gap traveling parallel to theplates at 2.0 x 1
Svetlanka [38]

Answer:

How far will the electron travel beforehitting a plate is 248.125mm

Explanation:

Applying Gauss' law:

Electric Field E = Charge density/epsilon nought

Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12

Therefore E = 1.0 x 10^-6/8.85× 10^-12

E= 1.13×10^5N/C

Force on electron F=qE

Where q=charge of electron=1.6×10^-19C

Therefore F=1.6×10^-19×1.13×10^5

F=1.808×10^-14N

Acceleration on electron a = Force/Mass

Where Mass of electron = 9.10938356 × 10^-31

Therefore a= 1.808×10^-14 /9.11 × 10-31

a= 1.985×10^16m/s^2

Time spent between plate = Distance/Speed

From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2

Therefore Time = 0.01/2×10^6

Time =5×10^-9s

How far the electron would travel S =ut+ at^2/2 where u=0

S= 1.985×10^16×(5×10^-9)^2/2

S=24.8125×10^-2m

S=248.125mm

4 0
2 years ago
A radio station's channel, such as 100.7 fm or 92.3 fm, is actually its frequency in megahertz (mhz), where 1mhz=106 hz and 1hz=
AURORKA [14]
The frequency of the radio station is
f=88.7 fm= 88.7 MHz = 88.7 \cdot 10^6 Hz

For radio waves (which are electromagnetic waves), the relationship between frequency f and wavelength \lambda is
\lambda= \frac{c}{f}
where c is the speed of light. Substituting the frequency of the radio station, we find the wavelength:
\lambda= \frac{3 \cdot 10^8 m/s}{88.7 \cdot 10^6 Hz}=3.38 m
6 0
3 years ago
Read 2 more answers
Nelle missioni di Apollo sulla Luna,il modulo di comando orbitava a un altitudine di 110 km al di sopra della superficie lunare.
liberstina [14]

Answer:

The period of orbit = 7143.41 s = 119.1 min = 1.984 hours

Il periodo dell'orbita = 7143.41 s = 119.1 min = 1.984 hours

Explanation:

English Translation

In Apollo's missions to the moon, the command module orbited at an altitude of 110 km above the lunar surface. How long did the command module complete an orbit?

Solution

According to Kepler's laws, the square of the period of orbit is proportional to the cube of the radius of orbit.

T² ∝ r³

T² = kr³

where k = constant of proportionality. The constant of proportionality is then given as

k = (4π²/GM)

It's all obtained from equating the gravitational force on the command module by the moon and the circular motion of the command module

Let

G = Gravitational constant

M = mass of the moon

m = mass of the command module

r = radius of orbit

w = angular velocity of the command module

(GMm/r²) = mrw²

(GM/r²) = rw²

(GM/r³) = w²

But angular velocity is given as

w = (2π/T)

w² = (4π²/T²)

(GM/r³) = (4π²/T²)

We then obtain

T² = (4π²/GM)r³

T² = kr³

Mass of moon = M = (7.35 x 10²²) kg Gravitational constant = G = (6.67 x 10⁻¹¹) Nm²/kg²

radius of moon = (1.74 x 10⁶) m

Total radius of orbit = 110 km + (radius of the moon) = 110,000 + (1.74 x 10⁶)

= (1.85 × 10⁶) m

k = (4π²/GM) = (4π² ÷ [(6.67 x 10⁻¹¹) × (7.35 x 10²²)] = (8.059 × 10⁻¹²) kg/Nm²

T² = kr³

T² = (8.059 × 10⁻¹²) × (1.85 × 10⁶)³

T² = 51,028,321.74

T = √(51,028,321.74) = 7,143.41107175

T = 7143.41 s = 119.1 min = 1.984 hours

Hope this Helps!!!

5 0
2 years ago
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