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Aleks04 [339]
3 years ago
15

The knights of labor do not believe in arbitration but believed in strikes? true or false

Physics
1 answer:
Rus_ich [418]3 years ago
5 0

Answer - The given statement is false.  

Explanation-

The members of knight of labour were opposing strikes and had a belief or boycott or arbitration.

A secret society was formed in 1869 by the tailors in Philadelphia and named Knights of labour. This union Organisation was formed to to get rights over equal pay if work is equal, abolish child labour, Work time of 8 hours and some graduated income tax  reforms. This union was vertically organized.

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a single closed path of electrical components including a voltage source

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What are the magnitude and direction of the acceleration of an electron at a point where the electric field has magnitude 6100 n
Hoochie [10]

Force on electron due to electric field is given by

F = eE

F = 1.6 * 10^{-19}* 6100

F = 9.76 * 10^{-16} N

now the acceleration is given by

a = \frac{F}{m}

a = \frac{9.76 * 10^{-16}}{9.1 * 10^{-31}}

a = 1.07 * 10^{15} m/s^2

so above is the magnitude of acceleration and its direction is opposite to field as electron is negatively charged so direction is towards SOUTH

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7. All electromagnetic waves travel at the same speed in empty space, the
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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
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