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maxonik [38]
2 years ago
10

I need help with this answer I believe it's a democracy​

Physics
1 answer:
kipiarov [429]2 years ago
8 0

Answer:

a. democracy

Explanation:

beacouse the government control of their members

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When electricity is produced from coal, the chemical energy in<br> coal iS changed to<br> energy.
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A point charge A of charge +4micro coloumb and another B of -1 micro coloumb are placed at a distance in air 1m apart then the d
andrew11 [14]

Answer:

Explanation:

Given that,

A point charge is placed between two charges

Q1 = 4 μC

Q2 = -1 μC

Distance between the two charges is 1m

We want to find the point when the electric field will be zero.

Electric field can be calculated using

E = kQ/r²

Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.

Then, the magnitude of the electric at point x is zero.

E = kQ1 / r² + kQ2 / r²

0 = kQ1 / x²  - kQ2 / (1-x)²

kQ1 / x² = kQ2 / (1-x)²

Divide through by k

Q1 / x² = Q2 / (1-x)²

4μ / x² = 1μ / (1 - x)²

Divide through by μ

4 / x² = 1 / (1-x)²

Cross multiply

4(1-x)² = x²

4(1-2x+x²) = x²

4 - 8x + 4x² = x²

4x² - 8x + 4 - x² = 0

3x² - 8x + 4 = 0

Check attachment for solution of quadratic equation

We found that,

x = 2m or x = ⅔m

So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.

5 0
3 years ago
The launching velocity of a missile is 20.0 m/s, and it is shot at 53º above the horizontal. What is the vertical component of t
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Stomach acid is a dilute solution of hydrochloric acid (HCl). In this case the word dilute means
vfiekz [6]
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3 years ago
Read 2 more answers
A 0.0550-kg ice cube at −30.0°C is placed in 0.400 kg of 35.0°C water in a very well-insulated container. What is the final temp
KatRina [158]

Answer:

19.34°C

Explanation:

When the ice cube is placed in the water, heat will be transferred from the hot water to it such that the heat gained (Q₁) by the ice is equal to the heat lost(Q₂) by the hot water and a final equilibrium temperature is reached between the melted ice and the cooling/cooled hot water. i.e

Q₁ = -Q₂                  ----------------------(i)

{A} Q₁ is the heat gained by the ice and it is given by the sum of ;

(i) the heat required to raise the temperature of the ice from -30°C to 0°C. This is given by [m₁ x c₁ x ΔT]

<em>Where;</em>

m₁ = mass of ice = 0.0550kg

c₁ = a constant called specific heat capacity of ice = 2108J/kg°C

ΔT₁ = change in the temperature of ice as it melts from -30°C to 0°C = [0 - (-30)]°C = [0 + 30]°C = 30°C

(ii) and the heat required to melt the ice completely - This is called the heat of fusion. This is given by [m₁ x L₁]

Where;

m₁ = mass of ice = 0.0550kg

L₁ = a constant called latent heat of fusion of ice = 334 x 10³J/kg

Therefore,

Q₁ = [m₁ x c₁ x ΔT₁] + [m₁ x L₁]        ------------------(ii)

Substitute the values of m₁, c₁, ΔT₁ and  L₁ into equation (ii) as follows;

Q₁ = [0.0550 x 2108 x 30] + [0.0550 x 334 x 10³]

Q₁ = [3478.2] + [18370]

Q₁ = 21848.2 J

{B} Q₂ is the heat lost by the hot water and is given by

Q₂ = m₂ x c₂ x ΔT₂                -----------------(iii)

Where;

m₂ = mass of water = 0.400kg

c₂ = a constant called specific heat capacity of water = 4200J/Kg°C

ΔT₂ = change in the temperature of water as it cools from 35°C to the final temperature of the hot water (T) = (T - 35)°C

Substitute these values into equation (iii) as follows;

Q₂ = 0.400 x 4200 x (T - 35)

Q₂ = 1680 x (T-35) J

{C} Now to get the final temperature, substitute the values of Q₁ and Q₂ into equation (i) as follows;

Q₁ = -Q₂

=> 21848.2 = - 1680 x (T-35)

=> 35 - T  = 21848.2 / 1680

=> 35 - T  = 13

=> T  = 35 - 13

=> T  = 22

Therefore the final temperature of the hot water is 22°C.

Now let's find the final temperature of the mixture.

The mixture contains hot water at 22°C and melted ice at 0°C

At this temperature, the heat (Q_{W}) due to the hot water will be equal to the negative of the one (Q_{I}) due to the melted ice.

i.e

Q_{W} = -Q_{I}             -----------------(a)

Where;

Q_{I} = m_{I} x c_{I} x ΔT_{I}         [m_{I} = mass of ice, c_{I} = specific heat capacity of melted ice which is now water and ΔT_{I} = change in temperature of the melted ice]

and

Q_{W} = m_{W} x c_{W} x ΔT_{W}    

[m_{W} = mass of water, c_{W} = specific heat capacity of water and ΔT_{W} = change in temperature of the water]

Substitute the values of Q_{W} and Q_{I} into equation (a) as follows

m_{W} x c_{W} x ΔT_{W}   =  - m_{I} x c_{I} x ΔT_{I}

Note that c_{W} and c_{I} are the same since they are both specific heat capacities of water. Therefore, the equation above becomes;

m_{W} x ΔT_{W}   = -m_{I} x ΔT_{I}   -----------------------(b)

Now, let's analyse ΔT_{W} and ΔT_{I}. The final temperature (T_{F}) of the two kinds of water(melted ice and cooled water) are now the same.

=> ΔT_{W} = change in temperature of water = final temperature of water(T_{F}) - initial temperature of water(T_{IW})

ΔT_{W} = T_{F} - T_{IW}

Where;

T_{IW} = 22°C           [which is the final temperature of water before mixture]

=> ΔT_{I} = change in temperature of melted ice = final temperature of water(T_{F}) - initial temperature of melted ice (T_{II})

ΔT_{I} = T_{F} - T_{II}

T_{II} = 0°C     (Initial temperature of the melted ice)

Substitute these values into equation (b) as follows;

m_{W} x ΔT_{W}   =  - m_{I} x ΔT_{I}

0.400 x (T_{F} - T_{IW}) = -0.0550 x (T_{F} - T_{II})

0.400 x (T_{F} - 22) = -0.0550 x (T_{F} - 0)

0.400 x (T_{F} - 22) = -0.0550 x (T_{F})

0.400T_{F} - 8.8 = -0.0550T_{F}

0.400T_{F} + 0.0550T_{F} =  8.8  

0.455T_{F} = 8.8

T_{F} = 19.34°C

Therefore, the final temperature of the mixture is 19.34°C

8 0
3 years ago
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