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Law Incorporation [45]

3 years ago

14

A point charge of +1.0 µC is moved in the direction of an electric field, and it has a change in electric potential difference e

nergy of 10.0 J. What was the change in electric potential difference? 1 µC = 10^-6 C
1.0 × 108 V

1.0 × 107 V

+1.0 × 107 V

+1.0 × 108 V,

1 answer:

olga55 [171]3 years ago

7 0

The answer of <span> the change in electric potential difference is</span> -1.0 × 10^7 V. It is defined as the work done per unit charge. A potential difference of 1 V means that 1 joule of work is done per coulomb of charge or 1 V = 1 J C-1.

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A satellite orbits earth with a mean altitude of 361 km. If the orbit is circular, what are the satellite's time period and spee

Advocard [28]

Answer:

v = 7.69 x 10³ m/s = 7690 m/s

T = 5500 s = 91.67 min = 1.53 h

Explanation:

In order for the satellite to orbit the earth, the force of gravitation on satellite must be equal to the centripetal force acting on it:

$F_{gravitation}= F_{centripetal}\\\\\frac{GM_{s} M_{E}}{r^2} = \frac{M_{s} v^2}{r}\\\\\frac{GM_{E}}{r} = v^2\\\\v = \sqrt{\frac{GM_{E}}{r} } \\\\$

where,

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Me = Mass of Earth = 5.97 x 10²⁴ kg

r = distance between the center of Earth and Satellite = Radius of Earth + Altitude = 6.371 x 10⁶ m + 0.361 x 10⁶ m = 6.732 x 10⁶ m

v = orbital speed = ?

Therefore,

$v = \sqrt{\frac{(6.67 x 10^{-11}N.m^2/kg^2)(5.97 x 10^{24} kg)}{6.732 x 10^6 m} }\\\\$

<u>v = 7.69 x 10³ m/s</u>

For time period satellite completes one revolution around the earth. It means that the distance covered by satellite is equal to circumference of circle at the given altitude.

So, its orbital speed can be given as:

$v = \frac{Circumference of Circle at Given Altitude}{T}\\\\v = \frac{2\pi r}{T}\\\\$

where,

T = Time Period of Satellite = ?

Therefore,

$T = \frac{2\pi r}{v}\\\\T = \frac{(2\pi )(6.732 x 10^6 m}{7.69 x 10^3 m/s}\\\\$

<u>T = 5500 s = 91.67 min = 1.53 h</u>

7 0

3 years ago

a particle is moving along a circular path having a radius of 4 in such that its position as a function of time is given by thet

ANTONII [103]

Answer:

Explanation:

Given

radius of circular path $r=4\ in.$

Position is given by

$\theta =\cos 2t---1$

Differentiate 1 to angular velocity we get

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega =-2\sin 2t----2$

Differentiate 2 to get angular acceleration

$\frac{\mathrm{d} \omega }{\mathrm{d} t}=-2^2\cos 2t ---3$

Net acceleration is the vector summation of tangential and centripetal force

$a_t=\alpha \times r$

$a_t=-4\cos 2t\times 4=-16\cos 2t$

$a_r=\omega ^2\cdot r$

$a_r=(-2\sin 2t)^2\cdot 4$

$a_r=16\sin^2(2t)$

$a_{net}=\sqrt{a_r^2+a_t^2}$

$a_{net}=\sqrt{(16\sin ^2(2t)+(-16\cos 2t)^2}$

$a_{net}=\sqrt{256\cos ^2(2t)+256\sin ^4(2t)}$

6 0

3 years ago

The what’s the difference does it mean velocity and vector Quantity is the same

ziro4ka [17]

Answer:

No they are totally different...

Velocity is displacement/time. while vector is both magnitude and direction.....

Velocity is a vector quantity because it has both magnitude and direction

6 0

3 years ago

I have two of these so if you think this is easy then another free ten points in my profile!

Agata [3.3K]

Answer:

z

Explanation:

x repersents a new moon and the others repersent quarter moons

(x is a new moon because new moons are often the phase when the moon is close to earths sun)

7 0

3 years ago

Read 2 more answers

A coil of wire containing N turns is in an external magnetic field that is perpendicular to the plane of the coil and it steadil

krok68 [10]

Answer:

The Resultant Induced Emf in coil is 4∈.

Explanation:

Given that,

A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.

To find :-

find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).

So,

Emf induced in the coil represented by formula

∈ = $-N\frac{d\phi}{dt}$ ...................(1)

Where:

. $\phi = BAcos\theta$ { B is magnetic field }

{A is cross-sectional area}

. $N =$ No. of turns in coil.

. $\frac{d\phi}{dt} =$ Rate change of induced Emf.

Here,

Considering the case :-

$N1 = 2N$ & $\frac{d\phi1}{dt} = 2\frac{d\phi}{dt}$

Putting these value in the equation (1) and finding the new emf induced (∈1)

∈1 = $-N1\times\frac{d\phi1}{dt}$

∈1 = $-2N\times2\frac{d\phi}{dt}$

∈1 = $4 [-N\times\frac{d\phi}{dt}]$

∈1 = 4∈ ...............{from Equation (1)}

Hence,

The Resultant Induced Emf in coil is 4∈.

8 0

3 years ago