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Sergio039 [100]
3 years ago
15

What form of braking is used to bring a vehicle to a smooth stop by applying smooth,steady pressure to the brake

Physics
1 answer:
Sav [38]3 years ago
7 0

Answer:

Controlled braking

Explanation:

CONTROLLED BRAKING occur in a situation where a person or an individual driving a vehicle releases the brake and slowly apply smooth as well as firmly pressure on the brake without the wheels been locked which is why CONTROLLED BRAKING are often used for emergency stops by drivers reason been that it helps to reduce speed when driving as fast as possible while the driver maintain the steering control of the vehicle.

Therefore the form of braking which is used to bring a vehicle to a smooth stop by applying smooth,steady pressure to the brake is called CONTROLLED BRAKING.

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The Kinetic energy, K, of an object with mass m moving with velocity v can be found using the formula - E_{\text{k}}={\tfrac {1}
tester [92]

Answer:

The ratio of kinetic energies of 5 kg object to 20 kg object is 1:1.

Explanation:

Kinetic energy is defined as energy possessed by an object due to its motion.It is calculated by:

K.E=\frac{1}{2}mv^2

Kinetic energy of the 5 kg object.

Mass of object,m = 5 kg

Velocity of an object = v

K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 5kg\times v^2

Kinetic energy of the 20 kg object.

Mass of object,m' = 20 kg

Velocity of an object = v'

K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 20kg\times v'^2

The ratio of the kinetic energy of the 5 kilogram object to the kinetic energy of the 20-kilogram object:

\frac{K.E}{K.E'}=\frac{\frac{1}{2}\times 5kg\times v^2}{\frac{1}{2}\times 20kg\times v'^2}

Given that, v = 2v'

\frac{K.E}{K.E'}=\frac{1}{1}

The ratio of kinetic energies of 5 kg object to 20 kg object is 1:1.

3 0
3 years ago
If light of wavelength 700 nm strikes such a photocathode, what will be the maximum kinetic energy, in eV , of the emitted elect
Oksana_A [137]

If the light of wavelength 700 nm strikes such a photocathode the maximum kinetic energy, in eV, of the emitted electrons is 0.558 eV.

so - $KE_{max} = hc/lembda}  work

threshold when KE = 0

hc/lambda = work = 1240/900=1.38 eV

b) Kemax = hc/lambda - work = 1240/640 -1.38=0.558 eV

What is photocathode?

  • A photocathode electrolyte interface can be used in a photoelectrolysis cell as the primary light-harvesting junction (in conjunction with an appropriate electrochemical anode) or as an optically complementary photoactive half-cell in a tandem photoelectrode photoelectrolysis cell (Hamnett, 1982; Kocha et al, 1994).
  • In the case of the former, the electrode should ideally harvest photon energy across the majority of the solar spectrum in order to achieve the highest energy conversion efficiency possible.
  • In the latter case, however, the photocathode may only be active in a specific band of the solar spectrum in order to generate a cathodic photocurrent sufficient to match the current generated in the photoanodic half-cell.

To learn more about Photocathode from the given link:

brainly.com/question/9861585

#SPJ4

3 0
2 years ago
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Is the force of the mosquito on the car larger than, smaller than, or equal to the force of the car on the mosquito?
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The mosquito exerts the same force on the car as the car exerts on the mosquito.
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