A 1800 kg hybrid vehicle operates on ethanol and is equipped with a multipurpose motorgenerator-flywheel. When the vehicle slows or stops, 50% of the kinetic energy is recovered as electrical energy in the battery. When the IC engine is used to recharge the battery, there is a 25% efficiency of converting chemical energy in the fuel to electrical energy stored in the battery. The vehicle slows from 70 miles per hour to 20 miles per hour. Calculate:

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A 1800 kg hybrid vehicle operates on ethanol and is equipped with a multipurpose motorgenerator-flywheel. When the vehicle slows or stops, 50% of the kinetic energy is recovered as electrical energy in the battery. When the IC engine is used to recharge the battery, there is a 25% efficiency of converting chemical energy in the fuel to electrical energy stored in the battery. The vehicle slows from 70 miles per hour to 20 miles per hour. Calculate: (A) Electrical energy recovered in the battery in [kJ] (B) Mass of fuel needed to store same amount of energy in the battery in [kg]

a) Electrical energy recovered in the battery is 404.6895 kJ

b) Mass of fuel needed to store same amount of energy in the battery is 0.0606 kg

Explanation:

Given that;

Initial speed of the vehicle V = 70 miles per hour = 31.293 m/s

Final speed of the vehicle u = 20 miles per hour = 8.941 m/s

mass of vehicle m = 1800 kg

Noe, change in kinetic energy of the vehicle will be;

= m( v² - u² )

we substitute

=   × 1800( (31.293)² - (8.941)² )

= 900( 979.2518 - 79.9414)

= 900 × 899.3104

=  809379.36 J

= 809.379 kJ

now, Electrical energy recovered in the battery when the vehicle slows will be;

= 50% ×

= 50/100 × 809.379 kJ

=  404.6895 kJ

Therefore, Electrical energy recovered in the battery is 404.6895 kJ

b)

For this electrical energy to be obtained from fuel, the chemical energy required will be;

=   / 25%

=  404.6895 kJ / 0.25

 = 1618.758 kJ  

Heat energy released per mass of ethanol combustion

(Lower heating value of ethanol) is 26.7kJ/g

Now, the mass of fuel needed to generate  1618.758 kJ will be;

= 1618.758 kJ / 26.7kJ/g

= 60.63 g

= 0.0606 kg

Therefore, Mass of fuel needed to store same amount of energy in the battery is 0.0606 kg

Independent variable: the best method to get rid of them.

Dependent variable: washing with soap and water.

Hypothesis: Organic oils

Control group: water

Answer:

Step-by-step explanation:

D.
Heat each cube to the same temperature, place each cube into different containers with 500 grams of water at the same temperature, and measure the temperature of the water.

Answer:

Step-by-step explanation:

To solve this problem, we can use the conservation of energy and conservation of momentum principles.

Conservation of energy:

The total initial energy is the rest energy of the proton and neutron, which is given by:

Ei = (mp + mn)c^2

where mp and mn are the masses of the proton and neutron, respectively, and c is the speed of light.

The total final energy is the rest energy of the deuteron plus the energy of the gamma ray, which is given by:

Ef = (md)c^2 + Eg

where md is the mass of the deuteron and Eg is the energy of the gamma ray.

According to the conservation of energy principle, the initial energy and final energy must be equal, so we have:

Ei = Ef

(mp + mn)c^2 = (md)c^2 + Eg

Conservation of momentum:

The total initial momentum is zero because the proton and neutron are at rest. The total final momentum is the momentum of the deuteron and the momentum of the gamma ray. Since the gamma ray is massless, its momentum is given by:

pg = Eg/c

where pg is the momentum of the gamma ray.

According to the conservation of momentum principle, the total final momentum must be equal to zero, so we have:

0 = pd + pg

where pd is the momentum of the deuteron.

Solving for md and pd:

From the conservation of energy equation, we can solve for md:

md = (mp + mn - Eg/c^2)/c^2

Substituting this expression into the conservation of momentum equation, we get:

pd = -pg = -Eg/c

Substituting the given values, we have:

mp = 1.6726 × 10^-27 kg mn = 1.6749 × 10^-27 kg Eg = 2.2 × 10^6 eV = 3.52 × 10^-13 J

Using c = 2.998 × 10^8 m/s, we get:

md = (1.6726 × 10^-27 kg + 1.6749 × 10^-27 kg - 3.52 × 10^-13 J/(2.998 × 10^8 m/s)^2)/(2.998 × 10^8 m/s)^2 = 3.3435 × 10^-27 kg

pd = -Eg/c = -(3.52 × 10^-13 J)/(2.998 × 10^8 m/s) = -1.1723 × 10^-21 kg·m/s

Therefore, the mass of the deuteron is 3.3435 × 10^-27 kg, and its momentum is -1.1723 × 10^-21 kg·m/s.

Answer:

Step-by-step explanation:

Time taken by the tomatoes to each the ground

using h = 1/2 g t^2 

t^2 = 2h/g = 2 x 50/ 9.8 = 10.2

t = 3.2 sec 

horizontal ditance = speed x time = 3 x 3.2 = 9.6 meters