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bagirrra123 [75]
3 years ago
11

A car manufacturer wants to change its car’s design to increase the car’s acceleration. Which changes should the engineers consi

der making to the design?
Physics
1 answer:
likoan [24]3 years ago
4 0

Answer:

The car manufacturers could increase bore of the cylinders, place the engine in the center or back of the car, add 1 to 2 turbochargers, and lower the center of gravity of the vehicle to increase traction.

Explanation:

Turbochargers would be recommended because they significantly increase both the torque of the engine as well as the amount of horses powering the car while also increasing original efficiency both with and without the additional power. Weight adjustment allows for lightweight vehicles with good traction. This is important to both keep control of the car under acceleration, but it also makes the vehicle more efficient due to the now sheddable unnecessary weight. A more obvious approach would be to increase the base horsepower and torque of the engine by increasing the bore of the cylinders and the weight of the pistons. This acts as an inertial lever, because the extra piston weight will drag the crankshaft faster. This could also be achieved by taking away piston weight, but this could be catastrophic should a piston slip.

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Which vector shows the direction of the centripetal acceleration at this point
Mademuasel [1]

Answer:

It's B

Explanation:

6 0
3 years ago
How large must the coefficient of static friction be between the tires and road, if a car rounds a level curve of radius 85 m at
tatuchka [14]

Answer:

0.66

Explanation:

By using the formula

u = v^2 / r g

Where u is coefficent of friction

u = 23.5 × 23.5 / (85 × 9.8)

u = 0.66

4 0
3 years ago
The kinetic energy of a rocket is increased by a factor of eight after its engines are fired, whereas its total mass is reduced
Rudik [331]

The momentum increases by a factor of 2

Explanation:

We can solve this problem by rewriting the momentum of the rocket in terms of the kinetic energy and the mass.

The kinetic energy of the rocket is:

K=\frac{1}{2}mv^2 (1)

where

m is the mass

v is the velocity

The momentum of the rocket is

p=mv (2)

From eq.(1) we get

v=\sqrt{\frac{2K}{m}}

and substituting into (2),

p=\sqrt{2mK}

Now in this problem we have:

- The kinetic energy of the rocket is increased by a factor 8:

K' = 8K

- The mass is reduced by half:

m'=\frac{m}{2}

Substituting, we find the new momentum:

p'=\sqrt{2(\frac{m}{2}(8K)}=\sqrt{4(2mK)}=2\sqrt{2mK}=2p

So, the momentum increases by a factor of 2.

Learn more about momentum and kinetic energy:

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#LearnwithBrainly

4 0
3 years ago
What is time write its SI unit​
lisabon 2012 [21]
The SI unit of time is second
4 0
3 years ago
Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars
zhuklara [117]

Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

T^2 \propto R^3

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

T^2 \propto R^3

T^2 = kR^3

Substituting the values, we get the value of constant k for mars.

687^2 = k\times (2.279 \times 10^{11})^3

k = 3.92 \times 10^{-29}

The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

365^3 = 3.92\times 10^{-29} R^3

R^3 = 3.39 \times 10^{33}

R= 1.50 \times 10^{11}\;\rm m

Hence we can conclude that the distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

To know more about Kepler's third law, follow the link given below.

brainly.com/question/7783290.

6 0
3 years ago
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