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romanna [79]
1 year ago
13

While practicing S-turns, a consistently smaller half-circle is made on one side of the road than on the other, and this turn is

not completed before crossing the road or reference line. This would most likely occur in turn:_______
Physics
1 answer:
erica [24]1 year ago
5 0

Answer:

The answer is "4-5-6 because the bank is growing too quickly in the beginning of its turn".

Explanation:

The S-turns is also known as the reference technique, in which the ground track of the aircraft on both sides of a defined ground-based straight-line distance represents 2 different but equivalent circles.  

Throughout the S-turns, on either side of the road, a progressively smaller half-circle is formed, and this turn does not stop until the road or reference line is crossed during the early part of the turn, the twists increase too quickly.

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A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along
storchak [24]

Answer:

v = \frac{kQ}{a}  

Explanation:

We define the linear density of charge as:

\lambda = \frac{Q}{L}

     Where L is the rod's length, in this case the semicircle's length L = πr

The potential created at the center by an differential element of charge is:

dv = \frac{kdq}{r}

          where k is the coulomb's constant

                     r is the distance from dq to center of the circle

Thus.

v = \int_{}^{}\frac{kdq}{a}  

v = \frac{k}{a}\int_{}^{}dq

v = \frac{kQ}{a}     Potential at the center of the semicircle

4 0
2 years ago
What is the threshold frequency for sodium metal if a photon with frequency 6.66 × 1014 s−1 ejects a photon with 7.74 × 10−20 J
FrozenT [24]

Answer:

5.5 × 10^14 Hz or s^-1

no orange light has less frequency so no photoelectric effect

Explanation:

hf = hf0 + K.E

HERE h is Planck 's constant having value 6.63 × 10 ^-34 J s

f is frequency of incident photon and f0 is threshold frequency

hf0 = hf- k.E

6.63 × 10 ^-34 × f0 = 6.63 × 10 ^-34× 6.66 × 10^14 - 7.74× 10^-20

6.63 × 10 ^-34 × f0 = 3.64158×10^-19

                           f0 = 3.64158×10^-19/ 6.63 × 10 ^-34

                           f0 = 5.4925 × 10^14

                            f0 =5.5 × 10^14 Hz or s^-1

frequency of orange light is 4.82 × 10^14 Hz which is less than threshold frequency hence photo electric effect will not be observed for orange light

8 0
2 years ago
Urban cities like Atlanta have to contend with a serious problem like pollution. Drivers in California are testing out a car tha
riadik2000 [5.3K]

1. With this statement, the author is referring to the fact that the vehicles are one of the largest polluters of the air. In order to reduce the pollution, the vehicles that are used will need to be changed, and with it the pollution will decrease significantly. The reduction of the pollution will come because the vehicles on hydrogen will not cause any pollution, so the enormous amounts of carbon dioxide released from the combustion of the engines will be thing of the past.

2. There are several challenges with this type of vehicles in order for them to replace the fossil fuel driven ones. The big price is one of the factors, as the majority of the people can not afford these cars. Another problem is that these vehicles are not as fast as the fossil fuel driven ones, and lot of people enjoy fast driving, despite it not being safe. There are millions of vehicles out there on the roads, and changing all of them with hydrogen vehicles will take a lot of time, as lot of those vehicles are new ones, so the people will not be willing to just throw them away and leave them rot in their garages. In order for the change of the driving park to be accomplished, the prices should go down, the people to be more serious about the environment and its protection, and patience as several decades will probably be needed for a change like this to be competed.

3. The fuel celled cars are a viable answer to decrease the pollution, as they are not causing any pollution, but instead will stop the process of large emissions of carbon dioxide from the fossil fueled cars. While this method is a good one, it should not be the only, as on its own it can not have the desirable effect, but instead all the major polluters should be included in the process. The industry and the production of energy are one of the major polluters as well, so they will need to follow the example, as if they not, the problem will stay, considering that the industry is constantly growing and the demand for energy is constantly growing too.

4 0
2 years ago
1 lb equals how many grams
shusha [124]

Answer:

1 lb. is 453.592 grams

Explanation:

1lb is 453.592 grams

8 0
2 years ago
Read 2 more answers
If the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The vol
xenn [34]

Answer:

Average density of Sun is 1.3927 \frac{g}{cm}.

Given:

Radius of Sun = 7.001 ×10^{5} km = 7.001 ×10^{10} cm

Mass of Sun = 2 × 10^{30} kg = 2 × 10^{33} g

To find:

Average density of Sun = ?

Formula used:

Density of Sun = \frac{Mass of Sun}{Volume of Sun}

Solution:

Density of Sun is given by,

Density of Sun = \frac{Mass of Sun}{Volume of Sun}

Volume of Sun = \frac{4}{3} \pi r^{3}

Volume of Sun = \frac{4}{3} \times 3.14 \times [7.001 \times 10^{10}]^{3}

Volume of Sun = 1.436 × 10^{33} cm^{3}

Density of Sun = \frac{ 2\times 10^{33} }{1.436 \times 10^{33} }

Density of Sun = 1.3927 \frac{g}{cm}

Thus, Average density of Sun is 1.3927 \frac{g}{cm}.

4 0
2 years ago
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