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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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how much power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m
Vesna [10]

2.89watts.

<h3>What is meant by sound intensity?</h3>
  • The average rate at which sound energy moves across a unit area normal to a given direction is used to determine a sound's intensity. This rate is generally stated in ergs per second per square centimeter.
  • Decibels are the units used to measure sound intensity, often known as sound power or sound pressure. The decibel (dB) unit is named after Alexander Graham Bell, who also created the audiometer and the telephone. An audiometer is a tool to gauge a person's hearing capacity for various noises.
  • Our ability to measure the flow of sound energy as a time-averaged vector quantity makes sound intensity measuring an effective method. We can identify sound sources and tell direct sound from reverberant sound in a room using the characteristics of sound intensity.

How much power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m:

Formula: I\frac{P}{4\pi r^{2} }

I=1.6x10-3 w/m2

r=12m

I\frac{P}{4\pi r^{2} }

P=I4\pi r^{2}

   =(1.6x10-3\frac{W}{m^{2} } ) (4\pi (12m)^{2} )

   =2.89watts.

To learn more about sound intensity, refer to:

brainly.com/question/17062836

#SPJ9

8 0
2 years ago
A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v
atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

y(x, t) = Acos(kx -wt)

The relationship between velocity and propagation constant is

v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\

v = 15.87m/s

Time taken, t = \frac{\lambda}{v}

= \frac{1.3}{15.87}\\\\=0.0819 sec

t = 0.0819s

2)

The velocity of transverse wave is given by

v = \sqrt{\frac{T}{\mu}}

v = \sqrt{\frac{W}{\frac{m}{\lambda}}}

mass of string is calculated thus

mg = 0.0125N

m = \frac{0.0125N}{9.8N/s}

m = 0.00128kg

\omega = \frac{v^2m}{\lambda}

\omega = \frac{(15.87^2)(0.00128)}{1.30}

\omega = 0.25N

3)

The propagation constant k is

k=\frac{2\pi}{\lambda}

hence

\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}

\lambda = 0.036 m

No of wavelengths, n is

n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\

n = 36

4)

The equation of wave travelling down the string is

y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]

without, unit\\\\y(x , t)= Acos[172x + 2730t]

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