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ankoles [38]
3 years ago
6

To understand the electric potential and electric field of a point charge in three dimensions Consider a positive point charge q

, located at the origin of three-dimensional space. Throughout this problem, use k in place of 14?
Physics
1 answer:
Liula [17]3 years ago
5 0

Answer:

The magnitude of the electric field at distance r from the point charge q is \dfrac{kq}{r^2}

Explanation:

Suppose, Find E(r), the magnitude of the electric field at distance r from the point charge q.

We know that,

The electric field is proportional to the point charge and inversely proportional to the square of the distance from the charge point .

In mathematically,

E=\dfrac{kq}{r^2}

Where, q = charge point

R = distance from charge point

Hence, The magnitude of the electric field at distance r from the point charge q is \dfrac{kq}{r^2}

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Breathe and now I’m just filling in more letters so it’ll go thru
3 0
3 years ago
A merchant in Katmandu sells you a solid gold 1.00-kg statue for a very reasonable price. When you get home, you wonder whether
White raven [17]

Answer:

a) 51.8 cm³

b) kg/m³ is a dimension of density (mass/volume). The regular unitys for volume are m³, cm³, L, gallons.

Explanation:

a) The density of pure gold is 19.3 g/cm³. When put in water, the piece of gold will occupy a volume, so that the volume of water will be displaced. To know the volume, we must divide the mass for the density (mass must be in grams because of the units of the density)

V = 1000/19.3

V = 51.8 cm³

7 0
3 years ago
A particle with charge −− 5.90 nCnC is moving in a uniform magnetic field B⃗ =−(B→=−( 1.28 TT )k^k^. The magnetic force on the p
maks197457 [2]

Answer:

Explanation:

Given that,

Charge q=-5.90nC

Magnetic field B= -1.28T k

And the magnetic force

F =−( 3.70×10−7N )i+( 7.60×10−7N )j

Let the velocity be V(xi + yj + zk)

Then, the force is given as

Note i×i=j×j×k×k=0

i×j=k. j×i=-k

j×k=i. k×j=-i

k×i=j. i×k=-j

The force in a magnetic field is given as

F= q(v×B)

−( 3.70×10−7N )i+( 7.60×10−7N )j =

q(xi + yj + zk) × -1.28k

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( -1.28x i×k - 1.28y j×k - 1.28z k×k)

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( 1.28xj - 1.28y i )

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( -1.28y i + 1.28x j)

So comparing comparing coefficients

let compare x axis component

-( 3.70×10−7N )i=-1.28qy i

−3.70×10−7N = -1.28qy

y= -3.7×10^-7/-1.28q

y= -3.7×10^-7/-1.28×-5.90×10^-9)

y=-48.99m/s

y≈-49m/s

Let compare y-axisaxis

7.6×10−7N j = 1.28qx j

7.6×10−7N = 1.28qx

x= 7.6×10^-7/1.28q

x= 7.6×10^-7/1.28×-5.90×10^-9

x=-100.64m/s

a. Then, the velocity of the x component is x= -100.64m/s

b. Also, the velocity component of the y axis is y=-49m/s

c. We will compute

V•F

V=-100.64i -49j

F=−( 3.70×10−7 N )i+( 7.60×10−7 N )j

Note

i.j=j.i=0. Also i.i = j.j =1

V•F is

(-100.64i-49j)•−(3.70×10−7N)i+(7.60×10−7 N )j =

3.724×10^-5 - 3.724×10^-5=0

V•F=0

d. Angle between V and F

V•F=|V||F|Cosx

0=|V||F|Cosx

Cosx=0

x= arccos(0)

x=90°

Since the dot product is zero, from vectors , if the dot product of two vectors is zero, then the vectors are perpendicular to each other

4 0
3 years ago
Un avión vuela a 10000m de altura y otro a 33300 pies, si un pie equivale a 30.48cm ¿Cuál vuela a mayor altura?
user100 [1]

Answer:

Avion A (10000 meters).

Explanation:

Deje que la altura de los aviones sea A y B respectivamente.

Dados los siguientes datos;

Altura A = 10000 metros

Altura B = 33300 pies

Para encontrar el avión que voló más alto, tendríamos que hacer alguna conversión de unidades.

Conversión:

Metros a centímetros;

1 metro = 100 cm

10000 metros = 100 * 10000 = 1.000.000 centímetros.

Por lo tanto, la altura A en cm = 1,000,000 centímetros

Pies a centímetros;

1 pie = 30,48 centímetros

33300 pies = 33300 * 30,48 = 1014984 centímetros.

Por lo tanto, la altura B en cm = 1014984 centímetros.

De los cálculos anteriores, podemos deducir que el avión A voló más alto.

6 0
3 years ago
A ball is released from a tower at a height of 100 meters toward the roof of another tower that is 25 meters high. The horizonta
I am Lyosha [343]
-- During the time the ball is flying from the high roof to the low roof,
it's going to fall (100-25) = 75 meters.
How long does it take an object dropped from rest to fall 75 meters ?

Distance = (1/2) · (gravity) · (time)²

75 m = (4.9 m/s²) · (time)²

Time² = (75 m) / (4.9 m/s²)
Time² = 15.31 sec²
Time = √(15.31 sec²)  =  3.91 seconds 

So the ball has to cover the horizontal distance of 20 meters 
in 3.91 seconds.

Distance = (speed) · (time)

20 m = (speed) · (3.91 sec)

Speed = (20 m) / (3.91 sec)

Speed  =  5.11 m/s
7 0
3 years ago
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