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3 years ago

How does drag affect the cost of owning a car?

Physics

2 answers:

vodka [1.7K]3 years ago

5 0

Answer:

You’ve probably noticed over the years that car design has become more streamlined. In other words, most consumer vehicles don’t look very boxy anymore. Some of this is due to aesthetics, but much of it is meant to decrease a vehicle’s drag coefficient. A cube has a high drag coefficient, whereas a teardrop has a low one.

By decreasing the drag coefficient, car makers are helping vehicles “slip” through air more easily. That reduces the amount of fuel needed to move the vehicle, and the difference shows up in your wallet and in the environment.

Explanation:

~Ban~

Elena L [17]3 years ago

4 0

Answer:

You move fast when moving only forward

Explanation:

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In the united states a standard letter sized piece of paper is 8.5 inches wide by 11 inches long. The international standard for

Andre45 [30]

This question is incomplete

Complete Question

In the United States, a standard letter-sized piece of paper is 8.5 inches wide by 11 inches long. The international standard for a letter-sized piece of paper is different. The international standard is based on SI units: 21.0 cm wide by 29.7 cm long.

a. Convert 21.0 cm to inches. Show your dimensional analysis setup.

b. Convert 29.7 cm to inches. Show your dimensional analysis setup.

c. State the dimensions, in inches, of the international standard for a letter-sized piece of paper.

d. Which piece of paper is longer: a U.S. letter-sized piece of paper, or an international letter-sized piece of paper?

Answer:

a) 8.267721 inches ≈ 8.3 inches

b) 11.6929197 inches ≈ 11.7 inches

c) It's dimensions in inches for the international standard for letter - sized for paper = 8.3 wide inches by 11.7 inches long

d) The International standard letter - sized paper is longer.

Explanation:

a. Convert 21.0 cm to inches. Show your dimensional analysis setup.

1 cm = 0.393701inch

21 cm =

Cross Multiply

21 cm × 0.393701inch/ 1 cm

= 8.267721 inches

Approximately 8.3 inches

b. Convert 29.7 cm to inches. Show your dimensional analysis setup.

1 cm = 0.393701inch

29.7 cm =

Cross Multiply

29.7 × 0.393701 inch/ 1 cm

= 11.6929197 inches

Approximately 11.7 inches

c. State the dimensions, in inches, of the international standard for a letter-sized piece of paper.

The international standard for a letter-sized has dimensions 21.0 cm wide by 29.7 cm long.

Where

21.0cm = 8.267721 inches

≈ 8.3 inches

29.7cm = 11.6929197 inches

≈ 11.7 inches

Hence, it's dimensions in inches = 8.3 inches by 11.7 inches.

d. Which piece of paper is longer: a U.S. letter-sized piece of paper, or an international letter-sized piece of paper?

U.S letter - sized paper = 8.5 inches wide by 11 inches long

International standard letter- sized paper = 8.3 wide inches by 11.7 inches long.

Hence, the International standard letter - sized paper is longer.

5 0

3 years ago

A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with clos

lukranit [14]

Answer:

The inductor contains $N = 523962.32$ loops

Explanation:

From the question we are told that

The capacitance of the capacitor is $C = 286nF = 286 * 10^{-9} \ F$

The resonance frequency is $f = 18.0 kHz = 18*10^{3} Hz$

The diameter is $d = 1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m$

The of the air-core inductor is $l = 12 \ m$

The permeability of free space is $\mu_o = 4 \pi *10^{-7} \ T \cdot m/A$

Generally the inductance of this air-core inductor is mathematically represented as

$L = \frac{\mu_o * N^2 \pi d^2}{4 l}$

This inductance can also be mathematically represented as

$L = \frac{1}{w^2}$

Where $w$ is the angular speed mathematically given as

$w = 2 \pi f$

$L = \frac{1}{4 \pi ^2 f^2}$

Now equating the both formulas for inductance

$\frac{\mu_o * N^2 \pi d^2}{4 l} = \frac{1}{4 \pi ^2 f^2}$

making N the subject of the formula

$N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C} }$

$N = \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }$

Substituting value

$N = \frac{1}{ 3.142 * 18*10^{3} * 0.00011 } \sqrt{\frac{12}{ 3.142 * 4 \pi *10^{-7}* 286 *10^{-9}} }$

$N = 523962.32$ loops

4 0

3 years ago

When there are more coils added to the wire of an electromagnet what happens to

fiasKO [112]

In technical terms, every coil of wire increases the "magnetic flux density" (strength) of your magnet.

So it's A (magnetic field increase)

5 0

3 years ago

What was Galileo’s contribution to the study of motion?

babunello [35]

Before giving you the answer straight away, I’d like to make sure you understand each option.

FIRST OPTION

“He developed the three laws of motion."

This option is not only the incorrect answer, but it is an incorrect statement. Galileo did not develop the three laws of motion. Isaac Newton was the one to compiled the three laws of motion.

SECOND OPTION

“He was the first to systematically study force and motion."

This option is a statement is true, and it could also be the correct answer. Galileo Galilei used experiments to search for the cause of motion.

THIRD OPTION

“He was the first to discover gravity."

This option is not only the incorrect answer, but it is an incorrect statement. Galileo was not the person to discover gravity. Sir Isaac Newton was the first to discover gravity.

FINAL OPTION

“He improved on Newton's laws."

This option is not only the incorrect answer, but it is an incorrect statement. Galilei did not improve Sir Isaac Newton’s laws.

Now that we’ve gone through each of the options individually, we can go ahead and determine the correct answer to the question, which is...

B ) He was the first to systematically study force and motion.

- Marlon Nunez

8 0

3 years ago

Read 2 more answers

A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th

ankoles [38]

<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Time, t = 3 s

Substituting

s = ut + 0.5 at²

s = 0 x 3 + 0.5 x 9.81 x 3²

s = 44.145 m

The seagull's approximate height above the ground at the time the clam was dropped is 4 m

4 0

4 years ago