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Pachacha [2.7K]
3 years ago
11

The energy expenditure value of traveling by car is 3.6 mj/passenger-kilometer. The value for traveling by train is 1.1 mj/passe

nger-kilometer. What would be the best way to increase the efficiency of traveling by car?
Physics
1 answer:
andrezito [222]3 years ago
7 0

Answer:

Using lighter material in car construction, improving energy efficiency by enhancing engine design or replacing the engine with more efficient technologies.

Explanation:

Using lighter materials in the car construction, reducing the potential energy required to accelerate and to move the car, as well as energy losses due to rolling friction. There is evidence of such benefits by replacing steel and aluminium parts with components made of composite materials.  

Improving the design of internal combustion engines to minimize energy losses and accordingly, improving energy efficiency. A more radical approach is replacing internal combustion engines with electric engines, which offer higher efficiencies. Such conclusions can be easily inferred from model based on Work-Energy Theorem and Principle of Energy Conservation:

\eta_{engine} \cdot U_{engine} = \frac{1}{2} \cdot m_{car} \cdot v^{2} + \mu_{r} \cdot m_{car} \cdot g \cdot \Delta s

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A "biconvex" lens is one in which both surfaces of the lens bulge outwards. Suppose you had a biconvex lens with radii of curvat
Stolb23 [73]

Answer: f=150cm in water and f=60cm in air.

Explanation: Focal length is a measurement of how strong light is converged or diverged by a system. To find the variable, it can be used the formula:

\frac{1}{f} = (nglass - ni)(\frac{1}{R1} - \frac{1}{R2}).

nglass is the index of refraction of the glass;

ni is the index of refraction of the medium you want, water in this case;

R1 is the curvature through which light enters the lens;

R2 is the curvature of the surface which it exits the lens;

Substituting and calculating for water (nwater = 1.3):

\frac{1}{f} = (1.5 - 1.3)(\frac{1}{10} - \frac{1}{15})

\frac{1}{f} = 0.2(\frac{1}{30})

f = \frac{30}{0.2} = 150

For air (nair = 1):

\frac{1}{f} = (1.5 - 1)(\frac{1}{10} - \frac{1}{15})

f = \frac{30}{0.5} = 60

In water, the focal length of the lens is f = 150cm.

In air, f = 60cm.

5 0
3 years ago
Read 2 more answers
What is the most abundant gas in the atmosphere?
Snowcat [4.5K]

Answer: Nitrogen

Explanation: Nitrogen makes up 78% of the atmosphere, Oxygen makes up 21%, and Argon 0.9%.

Water vaper makes up between 1-4%, depending on the region.

Carbon Dioxide makes up only about 0.04%.

8 0
3 years ago
How much heat is needed to boil 200g of water from 30 degrees:
tiny-mole [99]

Answer:

80 degress is needed to boil 200g of water from 30 degress

6 0
3 years ago
A 0.40-kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the stri
san4es73 [151]

Answer:

T = 8.55 N

Explanation:

When string makes an angle 40 degree with the vertical then it will have two forces on it

1) gravitational force (mg)

2) Tension force in string (T)

now we know that net force towards the center of the path is known as centripetal force and it is given as

T - mg cos40 = F_c

T - (0.40\times 9.8)cos40 = \frac{mv^2}{L}

T = 3 + \frac{0.40\times 5^2}{1.8}

T = 3 + 5.55

T = 8.55 N

6 0
3 years ago
According to Ohm’s law, the ratio of voltage to current in a conductor is
valina [46]

According to Ohm’s law, the ratio of voltage to current in a conductor is  constant.

Answer: Option A

<u>Explanation:</u>

Ohm's law defines that an electric current flowing through the conductor between two ends is directly proportionate to the voltage at these two points. The introduction of a constant proportionality, resistance, gives a simple mathematical equation describing this relationship. Particularly, Ohm's law also mentions that R is constant in this respect, i.e. not dependent on the current.

                                            I=\frac{V}{R}

Where,  

I is the current conduction in amperes

V is the voltage calculated by the conductor in volts

R is the conductor’s resistance in ohms.

3 0
3 years ago
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