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2 years ago

How does drag affect the cost of owning a car?

Physics

2 answers:

vodka [1.7K]2 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]2 years ago

4 0

Answer:

You move fast when moving only forward

Explanation:

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9. How much work is done when a 15kg box is lifted to a height of 2 meters?

torisob [31]

Answer: W = 294 J

Explanation: Solution:

Work is expressed as the product of force and the distance of the object.

W = Fd where F = mg

W= Fd

= mg d

= 15 kg ( 9.8 m/s²) ( 2m )

= 294 J

5 0

3 years ago

When sound travels through air, the particles _____. a. vibrate along the direction the wave travels b. vibrate but not in any f

Mandarinka [93]

Vibrate along the direction the wave travels

8 0

3 years ago

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0

2 years ago

A simple and common technique for accelerating electrons is shown in the figure, where there is a uniform electric field between

Sunny_sXe [5.5K]

Answer : $4.483\times 10^{15}\ m/s^2$ .

Explanation:

It is given that,

Electric field strength, $E=2.55\times 10^{4}\ N/C$

We know that,

Charge of electron, $q=1.6\times 10^{-19}\ C$

Mass of electron, $m=9.1\times 10^{-31}\ kg$

From the definition of electric field, $F=qE$ ...............(1)

According to Newton's second law, F = ma..........(2)

From equation (1) and (2)

$ma=qE$

$a=\dfrac{qE}{m}$

$a=\dfrac{1.6\times 10^{-19}\ C\times 2.55\times 10^{4}\ N/C }{9.1\times 10^{-31}\ kg}$

$a=0.4483\times 10^{16}\ m/s^2$

$a=4.483\times 10^{15}\ m/s^2$

So, the horizontal component of acceleration of an electron is $4.483\times 10^{15}\ m/s^2$ .

Hence, it is the required solution.

7 0

3 years ago

A 2-kg box sits on a horizontal table. the force of friction between the box and the table is 10 n. the box is pushed to the rig

dangina [55]

By Newton's 2nd law of motion, F = ma, where F is force, m is mass, and a is acceleration.

Rearranging this equation to find acceleration would give us:
a = F/m

The horizontal force to the right is 10N, because the box is pushed to the right with a force of 20N, and the friction force of 10N opposes that, so:
20N - 10N = 10N

The mass is 2kg.

Putting these values into the equation gives us:
a = F/m
= 10/2
= 5ms^-2

The acceleration of the box is 5ms^-2

6 0

3 years ago