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nasty-shy [4]
3 years ago
8

The election of 2008 had a record-breaking number of voters. It had the highest percentage of eligible voters voting in more tha

n
a decade. What is one thing that could have cause the 2008 election to have a higher turnout than 2000, 2004, and 2012?

People felt the need to vote out the incumbent.

People wanted to vote in a presidential election.

People wanted to keep the same political party in power.

People were particularly motivated by one of the candidates.

Physics
1 answer:
babunello [35]3 years ago
4 0

Answer:

People were particularly motivated by one of the candidates

Explanation:

The voter turnout in 2008 spiked because people were particularly motivated by Barrack Obama. His running for presidential position represented a <em>"generational change."</em> So, this means that many of the voters, especially the African-Americans ages 18 to 25 years old became interested in registering in order to vote.

Such kind of voter turnout spike is said to have also occurred during the candidacy of John F. Kennedy, where the voter turnout was<em> 62.8 percent.</em> This was because Kennedy was a <em>charismatic leader</em> who also represented a <u>generational change</u>.

So, this explains the answer.

You might be interested in
Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

4 0
3 years ago
A 30.0-kg girl in a swing is pushed to one side and held at rest by a horizontal force \vec{F} ​F ​⃗ ​​ so that the swing ropes
Virty [35]

Answer:

169.74 N

Explanation:

Given,

Mass of the girl = 30 Kg

angle of the rope with vertical, θ = 30°

equating the vertical component of the tension

vertical component of the tension is equal to the weight of the girl.

 T cos θ = m g

 T cos 30° = 30 x 9.8

 T = 339.48 N

Tension on the two ropes is equal to 339.48 N

Tension in each of the rope = T/2

                                           = 339.48/2 = 169.74 N

Hence, the tension in each of the rope is equal to 169.74 N

7 0
3 years ago
What is the difference between a mechanical wave and an electromagnetic wave
ivann1987 [24]

Answer:

1. Electromagnetic waves travel in a vacuum whereas mechanical waves do not.

2. The ripples made in a pool of water after a stone is thrown in the middle are an example of mechanical wave. Examples of electromagnetic waves include light and radio signals.

3. Mechanical waves are caused by wave amplitude and not by frequency. Electromagnetic Waves are produced by vibration of the charged particles.

4. While an electromagnetic wave is called just a disturbance, a mechanical wave is considered a periodic disturbance.

Explanation:

5 0
3 years ago
Two identical trucks have mass 5100 kg when empty, and the maximum permissible load for each is 8000 kg. the first truck, carryi
Oksanka [162]
<span>The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg. The key here is the conservation of momentum. For the first truck, the momentum is 0(5100 + 4300) The second truck has a starting momentum of 60(5100 + x) And finally, after the collision, the momentum of the whole system is 42(5100 + 4300 + 5100 + x) So let's set the equations for before and after the collision equal to each other. 0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x) And solve for x, first by adding the constant terms 0(5100 + 4300) + 60(5100 + x) = 42(14500 + x) Getting rid of the zero term 60(5100 + x) = 42(14500 + x) Distribute the 60 and the 42. 60*5100 + 60x = 42*14500 + 42x 306000 + 60x = 609000 + 42x Subtract 42x from both sides 306000 + 18x = 609000 Subtract 306000 from both sides 18x = 303000 And divide both sides by 18 x = 16833.33 So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums. 60(5100 + 16833.33) = 60(21933.33) = 1316000 42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000 They match. The 2nd truck was definitely over loaded.</span>
6 0
3 years ago
An insulated beaker with negligible mass contains 0.250 kg of water at 75.0C. How many kilograms of ice at -20.0C must be droppe
kkurt [141]

Answer:

The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg

Explanation:

Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water

Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C

To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.

Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C

Latent heat of ice = L = 334000 J/kg

Specific heat capacity of water = C = 4186 J/kg.°C

Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m

Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J

543600 m = 36627.5

m = 0.0674 kg = 67.4 g of ice.

3 0
3 years ago
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