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3 years ago

The driver and front seat passengers should sit at least _____ inches away from the air bag.

1 answer:

3 0

Air bags are kept in the steering wheel or dashboard and expand during a serious collision, usually a front collision that occurs at over 10 mph. To do its important job, an air bag comes out of the dashboard at up to 200 mph, faster than the blink of an eye. It is estimated around 10 inches of space to inflate. The force of an air bag can hurt those who are too close to it. Driver and front seat passengers should be moved as far back as practical, mainly people of short built. It is highly suggested that you be seated no less than 10 inches away from the air bag. Therefore, the answer is letter b.

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. Let us say that you put a cup of cold water in one room and a cup of hot water in

Lilit [14]

Answer:

In both cases, energy will move from an area of higher temperature to an area of lower temperature. So, the energy from room-temperature air will move into the cold water, which warms the water.

Explanation:

6 0

3 years ago

Read 2 more answers

An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7

Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

$v = \frac{I}{nqA}$

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

$n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}$

Now, we can find the drift speed:

$v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s$

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!

4 0

2 years ago

Someone help me haha;(

Nikitich [7]

Answer: vf = 51 m/s

d = 112 m

Explanation: Solution attached:

To find vf we use acceleration equation:

a = vf - vi / t

Derive to find vf

vf = at + vi

Substitute the values

vf = 3.5 m/s² ( 8.0 s) + 23 m/s

= 51 m/s

To solve for distance we use

d = (∆v)² / 2a

= (51 m/s - 23 m/s )² / 2 ( 3.5 m/s²)

= (28 m/s)² / 7 m/s²

= 784 m/s / 7 m/s²

= 112 m

6 0

3 years ago

If all the stars in an elliptical galaxy traveled random directions in their orbits, the elliptical galaxy would be type

Lemur [1.5K]

The answer would be E7. Galaxies categorized as E0 look to be nearly perfect, while those registered as E7 seem much extended than they are widespread. It is worth noting, though, that a galaxy's look is connected to how it lies on the sky when viewed from Earth. An E7 galaxy is very long and thin or the flattest of them all.

7 0

3 years ago

The metal gold crystallizes in a face centered cubic unit cell with one atom per lattice point. When X-rays with λ = 1.436 Å are

Umnica [9.8K]

Answer:

r = 1.45 Å

Explanation:

given,

λ = 1.436 Å

θ = 20.62°

d = a

n = 2

metal gold crystallizes in a face centered cubic unit cell

Radius of the gold atom = ?

using Bragg's Law

n λ = 2 d sin θ

2 x 1.436 Å = 2 a sin 20.62°

a = 4.077 Å

We know relation of radius for face centered cubic unit cell

$a = \dfrac{4r}{\sqrt{2}}$

$4.077= \dfrac{4\times r}{\sqrt{2}}$

r = 1.45 Å

the radius of a(n) gold atom. is equal to 1.45 Å

7 0

3 years ago