1.Which mechanical part or feature listed in the section on Critical Vehicle Systems do you think is most important?

The winning time for a 500.0 mile circular race track was 3 hours and

lyudmila [28]

Explanation:

We have,

Distance traveled in a circular track is 500 miles

The winning time was 3 hours and 13 minutes. It means time is 3.217 hours.

The driver's average speed is given by total distance divided by total time taken. Its formula can be written as :

$v=\dfrac{d}{t}\\\\v=\dfrac{500\ miles}{3.217\ h}\\\\v=155.42\ mph$

At the end of the race, the driver reaches the point form where he has started. It means the displacement of the driver is equal to 0. Hence, driver's average velocity is equal to 0.

3 0

4 years ago

A ray of light is incident on an air/water interface.

elixir [45]

Answer:

$\theta_2=sin^{-1}(\dfrac{n_1\ sin\theta_1}{n_2})$

Explanation:

Given that,

The ray makes an angle of 32 degrees with respect to the normal of the surface.

The refractive index of air, $n_1=1$

The refractive index of water, $n_2=1.33$

Snell's law is given by :

$n_1\ sin\theta_1=n_2\ sin\theta_2$

$sin\theta_2=\dfrac{n_1\ sin\theta_1}{n_2}$

$\theta_2=sin^{-1}(\dfrac{n_1\ sin\theta_1}{n_2})$

So, option (4) is correct. Hence, this is the required solution.

3 0

3 years ago

A mixture consists of sand, salt, and iron filings. Which of the following instruments or materials would not be useful in separ

Zanzabum

Well water would prob be your best bet, you could extract the salt first and then use a magnet to seperate the iron filing from the sand

5 0

4 years ago

P wave for 100km time= distance/speed , T=100km/6.1km/s

shutvik [7]

I dont know what you are asking sorry can u elaborate

8 0

3 years ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the

alex41 [277]

Answer:

(a) $v_1 = a_1t_1 = 1.76 t_1$

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

$v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1$

(b) The velocity of the car before the driver begins braking is

$v_1 = 1.76*20 = 35.2m/s$

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

$a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2$

We can use the following equation of motion to calculate how far the car has travel since braking to stop

$s_2 = v_1t_2 + a_2t_2^2/2$

$s_2 = 35.2*5 - 7.04*5^2/2 = 88 m$

Also the distance from start to where the driver starts braking is

$s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352$

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

8 0

3 years ago