All categories

3 years ago

Fog rolls in and bright headlights make it hard to see in the fog

Physics

1 answer:

Taya2010 [7]3 years ago

3 0

Answer:F., D.

Explanation:If you use low beams it should help and if you use the defogger it should also help.

You might be interested in

Calculate the unit cell edge length for an 79 wt% Ag- 21 wt% Pd alloy. All of the palladium is in solid solution, and the crysta

ankoles [38]

Answer:

The edge length is 0.4036 nm

Solution:

As per the question:

Density of Ag, $\rho = 10.49 g/cm^{3}$

Density of Pd, $\rho = 12.02 g/cm^{3}$

Atomic weight of Ag, A = 107.87 g/mol

Atomic weight of Pd, A' = 106.4 g/mol

Now,

The average density, $\rho_{a} = \frac{n A_{avg}} {V_{c}\times N_{A}}$

where

$V_{c} = a^{3}$ = Volume of crystal lattice

a = edge length

n = 4 = no. of atoms in FCC

Therefore,

$\rho_{a} = = \frac{n A_{avg}} {V_{c}\times N_{A}}$

Therefore, the length of the unit cell is given as:

$a = (\frac{nA_{avg}}{\rho_{a}\times N_{a}})^{1/3}$ (1)

Average atomic weight is given as:

$A_{avg} = \frac{100}{\frac{C_{Ag}}{A_{Ag}} + \frac{C_{Pd}}{A_{Pd}}}$

where

$C_{Ag}$ = 79 %

$A_{Ag}$ = 107

$C_{Pd}$ = 21%

$A_{Pd}$ = 106

Therefore,

$A_{avg} = \frac{100}{\frac{79}{107} + \frac{21}{106}} = 106.78$

In the similar way, average density is given as:

$\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}$

$\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}$

Therefore, edge length is given by eqn (1) as:

$a = (\frac{4\times 106.78}{10.78\times 6.023 X 10^23})^{1/3} = 4.036\times 10^{- 8} cm = 0.4036\times 10^{- 9} m = 0.4036 nm$

5 0

3 years ago

The dwarf planet Ceres contains over 50% of the mass of the main asteroid belt.

PolarNik [594]

False

Explanation:

Called an asteroid for many years, Ceres is so much bigger and so different from its rocky neighbors that scientists classified it as a dwarf planet in 2006. Even though Ceres comprises 25 percent of the asteroid belt's total mass, tiny Pluto is still 14 times more massive.

6 0

2 years ago

A student is watching waves come in from the ocean. He noticed that the first wave he saw (Wave A) had twice the amplitude of th

Alika [10]

Answer:

Wave A

I hope this helps

8 0

2 years ago

A 5000 kg open train car is rolling on frictionless rails at 22 m/s when it starts pouring rain. A few minutes later, the car’s

Phantasy [73]

Answer:

500 kg

Explanation:

It is given that,

The mass of a open train car, M = 5000 kg

Speed of open train car, V = 22 m/s

A few minutes later, the car’s speed is 20 m/s

We need to find the mass of water collected in the car. It is based on the conservation of momentum as follows :

initial momentum = final momentum

Let m is final mass

MV=mv

$m=\dfrac{MV}{v}\\\\m=\dfrac{5000\times 22}{20}\\\\=5500\ kg$

Water collected = After mass of train - before mass of train

= 5500 - 5000

= 500 kg

So, 500 kg of water has collected in the car.

3 0

3 years ago

HELP!!

raketka [301]

Hello!

Since the two weights are off the table, the block will move towards letter F.

I hope this helps :))

5 0

3 years ago

Read 2 more answers