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ddd [48]
3 years ago
8

How does a smoother surface affect braking distance

Physics
1 answer:
WARRIOR [948]3 years ago
3 0
Friction has to do with this.  If you have a smoother surface than another surface, the smoother surface will contain less friction.  Friction is an opposing force against a moving object.  If there is less opposing force towards a moving object, there is less friction.  If you have a smooth road, cars won't break as fast as they would break on bumpy roads, because bumpy roads have more bumps, creating more friction.  I hope that helped.  :D
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A rocket is fired at 100 m/s at an angle of 37, how many seconds is it in the air?
Sonbull [250]

Answer:

12.3 seconds

Explanation:

time in air = 2(v)/g

2(100sin(37))/9.8

=12.28 seconds

7 0
3 years ago
Hecto the Mornar a fursa explain i magnituse of the force acting right angle to the moment arm​
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<h2>~<u>Solution</u> :-</h2>
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3 years ago
Explain in details the concept of economic geography​
vova2212 [387]

Answer:

In the words of Hartshorn and Alexander: “Economic Geography is the study of the spatial variation on the earth’s surface of activities related to producing, exchanging and consuming goods and services. Whenever possible the goal is to develop generalizations and theories to account for these spatial variations.”

Explanation:

3 0
3 years ago
The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum
Tanya [424]

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case  h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W;  we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV

Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this  acts to slow down the ejected electrons from the catode.

5 0
3 years ago
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Soloha48 [4]
I think it is a. all matter

6 0
3 years ago
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