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Fudgin [204]
3 years ago
9

(i) Candidates and parties with a lot of money may not be sure of their victory but they do enjoy a big and unfair advantage ove

r smaller parties and independents. (ii) In some parts of the country, candidates with criminal connection have been able to push others out of the electoral race and to secure a ticket from major parties. (iii) Some families tend to dominate political parties. Tickets are distributed to relatives from these families.
Physics
1 answer:
madam [21]3 years ago
7 0

Answer:

Challenges to a fair election       

Explanation:

All the point mention or given in the question are challenges to a fair and free election process.

Election may be defined as an procedure to select or elect any person as a member of an organization or as an official. Here people cast their vote to their favorite person whom they want to choose and elect as their official. Though election is a just and fair process but there are several challenges to it.

It is seen that parties with more money try to influence their voters. Some candidates may have contact with the criminals and they use these contacts to play unfair means and try to win the election. While there are some families which have a dominance in the political parties and they try to win their family members in the election.  

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A domestic water heater holds 189 L of water at 608C, 1 atm. Determine the exergy of the hot water, in kJ. To what elevation, in
Gekata [30.6K]

A.

The energy of the hot water is 482630400 J

Using Q = mcΔT where Q = energy of hot water, m = mass of water = ρV where ρ = density of water = 1000 kg/m³ and V = volume of water = 189 L = 0.189 m³,

c = specific heat capacity of water = 4200 J/kg-°C and ΔT = temperature change of water = T₂ - T₁ where T₂ = final temperature of water = 608 °C. If we assume the water was initially at 0°C, T₁ = 0 °C. So, the temperature change ΔT = 608 °C - 0 °C = 608 °C

Substituting the values of the variables into the  equation, we have

Q = mcΔT

Q = ρVcΔT

Q = 1000 kg/m³ × 0.189 m³ × 4200 J/kg-°C × 608 °C

Q = 482630400 J

So, the energy of the hot water is 482630400 J

B.

The elevation <u>the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water</u> is 49248 m.

Using the equation for gravitational potential energy ΔU = mgΔh where m = mass of object = 1000 kg, g = acceleration due to gravity = 9.8 m/s² and Δh = h - h' where h = required elevation and h' = zero level elevation = 0 m

Since the energy of the mass equal the energy of the hot water, ΔU = 482630400 J

So, ΔU = mgΔh

ΔU = mg(h - h')

making h subject of the formula, we have

h = h' + ΔU/mg

Substituting the values of the variables into the equation, we have

h = h' + ΔU/mg

h = 0 m + 482630400 J/(1000 kg × 9.8 m/s²)

h = 0 m + 482630400 J/(9800 kgm/s²)

h = 0 m + 49248 m

h = 49248 m

So, the elevation <u>the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water</u> is 49248 m.

Learn more about heat energy here:

brainly.com/question/11961649

5 0
2 years ago
Find the magnitude of the sum of these two vectors: B 101 m 60.0 ° 85.0 m A​
slamgirl [31]

Answer:

M= 161.1 m Deg= 32.9

Explanation:

3 0
1 year ago
3. A cart with mass of 30 kg is traveling with a velocity of 4.0 m/s. The
MaRussiya [10]

Answer:

39 m/s

Explanation:

8 0
2 years ago
Read 2 more answers
2. the dipole moment of a dipole in a 300-n/c electric field is initially perpendicular to the field, but it rotates so it is in
aleksandr82 [10.1K]

Work Done (W) by the field is-6x 107 J,

<h3>What is Electric dipole?</h3>

A pair of opposite, non-coplanar, equally powerful electric charges that are in opposition to one another. An atom is said to have a "induced electric dipole" if the center of the negative cloud of electrons has moved a little bit away from the nucleus due to an external electric field. When the external field is taken away, dipolarity is lost.

Electric field (E) = 300 N/C

Dipole moment (p) = 2 x 10° Cm

Solution:

From the formula we know.

U = -pE cosФ

Here,

p Denotes Dipole moment.

E Denotes Electric field.

Ф Denotes angle b/w them

Now, as given, firstly the dipole is perpendicular to the electric field, so

angle (Ф1) will be 90° and now the dipole is rotated such that they are in same

direction so the angle (Ф2) will be 0°

So, let's find Change in Potential energy which will be equal to the work done

by the electric field.

ΔU = Uf - Ui

ΔU = [-pE*cos Ф2] - [-pE *cos Ф1]

ΔU = [-pE*cos Ф2] + pE *cos Ф1

ΔU = pE * [cos Ф2+ cos Ф1]

Substituting the values,

ΔU = pE * [cos 0° + cos 90°]

ΔU = pE * (-1 +0)

ΔU = -pE

ΔU = -2x 10^-9 × 300

ΔU = 6 x 10^(-9+2)

ΔU = 6 x 10^-7

W = ΔU = -6 x 10^-7

W = - 6 x 10 7 J

Work Done (W) by the field is - 6 x 10-7 J.

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6 0
1 year ago
A ball is thrown straight up at time t=0 with an initial speed of 19m/s. Take the point of release to be y0=0 and upwards to be
Wittaler [7]
First we write the corresponding kinematics equations:
 a = -g
 v = -g * t + vo
 y = -g * ((t ^ 2) / 2) + vo * t + yo
 Substituting the values:
 y = - (9.81) * (((0.50) ^ 2) / 2) + (19) * (0.50) + (0) = 8.27m
 answer:
 the displacement at the time of 0.50s is 8.27m
4 0
3 years ago
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