Ask question

Ask question

All categories

4 years ago

14

What is the proper hand position when driving straight ahead

1 answer:

vekshin14 years ago

7 0

NHTSA (National Highway Traffic Safety Administration) now recommends the technique known as 9 and 3. Place your left hand on the left portion of the steering wheel in a location approximate to where the nine would be if the wheel was a clock. Your right hand should be placed on the right portion of the wheel where the three would be located.

You might be interested in

What do all electromagnetic waves have in common?

klasskru [66]

Answer:
They have the same wavelength

5 0

4 years ago

What is the most effective means of establishing awareness of hazards in commercial, industrial, and storage facilities with lar

coldgirl [10]

Answer:

C: Contacting the facilities.

7 0

3 years ago

I'm thinking A?

Luba_88 [7]

Hey girl

the answer is A

your correct the atmosphere provides warmth

hope this helps

XD

5 0

3 years ago

19. A mass of gas has a volume of 4 m3, a temperature of 290 K, and an absolute pressure of 475 kPa. When the gas is allowed to

Aleks [24]

Recall this gas law:

$\frac{P₁V₁}{T₁}$ = $\frac{P₂V}{T₂}$

P₁ and P₂ are the initial and final pressures.

V₁ and V₂ are the initial and final volumes.

T₁ and T₂ are the initial and final temperatures.

Given values:

P₁ = 475kPa

V₁ = 4m³, V₂ = 6.5m³

T₁ = 290K, T₂ = 277K

Substitute the terms in the equation with the given values and solve for Pf:

$\frac{475*4}{290} = \frac{P₂*6.5}{277}$

<h3>P₂ = 279.2kPa</h3>

8 0

4 years ago

What visible wavelengths of light are strongly reflected from a 390-nm-thick soap bubble?

DiKsa [7]

Answer:

So visible wavelength which is possible here is

416 nm and 693.3 nm

Explanation:

As we know that for normal incidence of light the path difference of the reflected ray is given as

$2\mu t + \frac{\lambda}{2} = \Delta x$

so here we can say that for maximum intensity condition we will have

$\Delta x = N\lambda$

so we have

$2\mu t + \frac{\lambda}{2} = N\lambda$

now for visible wavelength we have

for N = 1

$2\mu t = \frac{\lambda}{2}$

$\lambda = 4\mu t$

$\lambda = 4(\frac{4}{3})(390 nm)$

$\lambda = 2080 nm$

for N = 2

$\lambda = \frac{4\mu t}{3}$

$\lambda = \frac{4(\frac{4}{3})(390 nm)}{3}$

$\lambda = 693.3 nm$

for N = 3

$\lambda = \frac{4\mu t}{5}$

$\lambda = \frac{4(\frac{4}{3})(390 nm)}{5}$

$\lambda = 416 nm$

6 0

4 years ago