Place the following into scientific notation...(.000635)<br> *

What type of electricity comes from an outlet

zalisa [80]

There’s many types of electricity that comes from an outlet, like [15A, 120 Volt Outlets] those are more common in older homes and can come in two versions [Two-pronged outlet and three-pronged version]

8 0

3 years ago

e force acting between two charged particles A and B is 5.2 × 10-5 newtons. Charges A and B are 2.4 × 10-2 meters apart. If the

anastassius [24]

The force acting between the particles is

$F=k \frac{Q_{1}Q_{2}}{r^2}$
Then
$Q_{2}= \frac{5.2 \times 10^-^5 \times 0.024^2}{ 9.0 \times 10^9 7.2 \times 10^-^8} =4.622 \times 10^-^1^1C$

7 0

2 years ago

Read 2 more answers

a) An electric circuit is rates, 240V, 13Ω calculate the amount of current that can be used by this appliance.

zaharov [31]

Answer:

18.4615385 amps

Explanation:

The voltage V in volts (V) is equal to the current I in amps (A) times the resistance R in ohms (Ω):

5 0

3 years ago

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su

yKpoI14uk [10]

Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

0 V/m

E⃗ 2

0 V/m

E⃗ 3

$E_3 = 2.153 *10^{9} \ V/m$

E⃗ 4

$E_4 = 5.754 *10^ {8} \ V/m$

Explanation:

From the question we are told that

The diameter is $d = 6.53 \mu m = 6.53*10^{-6}\ m$

The charge is $Q = -.2.55 *10^{-12} \ C$

The permittivity of free space is $\epsilon_o = 8.85* 10^{-12}\ C / V.m$

The distance considered is $d = 3.05 \mu m = 3.05 *10^{-6} \ m$

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

$E_3 = \frac{ k * |Q|}{ r^2 }$

Here $k$ is the coulomb constant with value

$k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}$

r is the radius of the sphere which is mathematically as

$r = \frac{d}{2} = \frac{6.53*10^{-6}}{2} = 3.265 *10^{-6} \ m$

$E_3 = \frac{ 9*10^{9} * |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }$

$E_3 = 2.153 *10^{9} \ V/m$

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

$E_4 = \frac{ k * |Q|}{ R^2 }$

Here R is mathematically represented as

$R = 3.265 *10^{-6} + 3.05 *10^{-6}$

=> $R = 6.315 *10^{-6}$

So

$E_4 = \frac{ 9*10^{9} * |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }$

$E_4 = 5.754 *10^ {8} \ V/m$

3 0

2 years ago

The turnbuckle is tightened until the tension in the cable AB equals 2.3 kN. Determine the vector expression for the tension T a

brilliants [131]

Answer:

a) 0.83984 i + 0.41992 j - 2.0996 k KN

b) T_ac = 1.972888 KN

Explanation:

Given:

- The tension in cable AB = 2.3 KN

Find:

a) Determine the vector expression for the tension T as a force acting on member AD.

b) Also find the magnitude of the projection of T along the line AC.

Solution:

part a)

- Find unit vector AB:

vector (AB) = 2 i + j - 5 k

mag (AB) = sqrt (2^2 + 1^2 + 5^2)

mag (AB) = sqrt(30)

unit (AB) = ( 1 / sqrt(30) )* ( 2 i + j - 5 k )

- Find Tension vector:

vector (T) = unit(AB)* 2.3 KN

= 0.83984 i + 0.41992 j - 2.0996 k

- The projection of T onto AC can be found from the dot product of vector T to unit vector (AC)

- For unit vector (AC)

vector (AC) = 2 i - 2 j - 5 k

mag (AC) = sqrt (2^2 + 2^2 + 5^2)

mag (AC) = sqrt(33)

unit (AC) = ( 1 / sqrt(33) )* ( 2 i - 2 j - 5 k )

- Compute the projection:

T_ac = vector T . unit (AC)

T_ac = (0.83984 i + 0.41992 j - 2.0996 k) . ( 1 / sqrt(33) )* ( 2 i - 2 j - 5 k )

T_ac = 0.2923947572 - 0.146973786 - 1.827467232

T_ac = 1.972888 KN

4 0

3 years ago

Place the following into scientific notation...(.000635) *