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NemiM [27]
3 years ago
5

The magnitude E of an electric field depends on the radial distance r according to E = A/r4, where A is a constant with unit vol

t-cubic meter. As a multiple of A, what is the magnitude of the electric potential difference between r1 = 1.71 m and r2 = 2.89 m?
Physics
1 answer:
Lesechka [4]3 years ago
8 0

Answer:

\Delta V = 0.053 A

Explanation:

Electric field in a given region is given by equation

E = \frac{A}{r^4}

as we know the relation between electric field and potential difference is given as

\Delta V = -\int E. dr

so here we have

\Delta V = - \int (\frac{A}{r^4}) .dr

\Delta V = \frac{A}{3r_1^3} - \frac{A}{3r_2^3}

here we know that

r_1 = 1.71 m  and r_2 = 2.89 m

so we will have

\Delta V = \frac{A}{3}(\frac{1}{1.71^3} - \frac{1}{2.89^3})

so we will have

\Delta V = 0.053 A

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What is the total wavelength if one-half of the wave is 3?
seraphim [82]

Answer:

6

Explanation:

Half the wave = 3

Wavelength = 3 x 2 = 6

3 0
2 years ago
(b) A piece of wood of volume 0.6 m² floats in water. Find the volume
enot [183]

Answer:

Explanation:

Let the volume below water be v . Then

buoyant force = v d g where d is density of water , g is acceleration due to gravity

= v x 1000 x g

weight of wood piece = volume x density of wood x g

= .6 x 600 x g

for equilibrium while floating

buoyant force = weight

= v x 1000 x g  =  .6 x 600 x g

v = .36 m²

volume above water or volume exposed = .6 - .36

= .24 m²

When immersed completely ,

buoyant force = .6 x 1000 x 9.8

= 5880 N

weight of wood

=  .6 x 600 x g

= 3528 N

buoyant force is more than the weight . In order to equalise them for floating with full volume in water

weight required = 5880 - 3528

= 2352 N.

6 0
3 years ago
An office building has a 24-volt branch circuit installed for landscape lighting around the front of the building. The circuit w
Arturiano [62]

The circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.

<h3>UF cable</h3>

UF cable is used as an underground feeder cable to distribute power from an existing building to outdoor equipment. UF cable can also be used as direct burial cable.

For the 24-volt branch circuit installed, the minimum burial depth will be 6 inches.

Thus, the circuit was installed in UF cable which requires a minimum burial depth of 6 inches for this circuit.

Learn more about UF cable here: brainly.com/question/8591560

5 0
2 years ago
Se necesita subir una carga de 500 kg (4900 N) a una altura de 1.5 m deslizándola sobre una rampa inclinada. ¿Qué longitud debe
marusya05 [52]

Answer:

4.22 m

Explanation:

Una rampa es una máquina que se utiliza para levantar un objeto con una fuerza menor a la que realmente necesitarías. Cuanto mayor sea la longitud de la rampa, menor será la magnitud de la fuerza necesaria para levantar el objeto.

Dado que:

altura de la rampa = 1.5 m, carga = 4900 N, fuerza aplicada = 1633.33 N.

La fórmula de la rampa se da como:

fuerza aplicada * longitud de la rampa = peso de la carga * altura de la rampa

1633.33 * longitud de la rampa = 4900 * 1.5

longitud de la rampa = 4900 * 1.5 / 1633.33

longitud de la rampa = 4.22 m

6 0
3 years ago
In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere
muminat

Answer:

2.1\cdot 10^{21} electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

E=\frac{kQ}{r^2}

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

In this problem, the cloud is assumed to be a  charged sphere, so we have:

E_b=3.00\cdot 10^6 N/C is the maximum electric field strength tolerated by the air before breakdown occurs

r=1.00 km = 1000 m is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C

Assuming that the cloud is negatively charged, then

Q=-333.3 C

And since the charge of one electron is

e=-1.6\cdot 10^{-19}C

The number of excess electrons on the cloud is

N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}

5 0
2 years ago
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