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PIT_PIT [208]
3 years ago
9

Suppose that a 1000 kg car is traveling at 25 m/s. Its brakes can apply a force of 5000N. What is the minimum distance required

for the car to stop?
Physics
1 answer:
stealth61 [152]3 years ago
6 0
The car's kinetic energy is

                             (1/2) (mass) (speed)²

                       =    (1/2) (1,000 kg) (25 m/s)²

                       =      (500 kg)  (625 m²/s²)

                       =        312,500 joules .

THAT's the work the brakes have to do in order to stop the car.
They have to absorb that kinetic energy and send it somewhere.

Work done by the brakes = (force) x (distance)

             312,500 joules  =  (5,000 N) x (distance)

                     Distance  =  (312,500 joules) / (5,000 N)

                                   =     62.5 meters .

The brakes soak up the car's kinetic energy, turn it to heat,
and let it blow away in the wind.

Too bad you paid good money to buy that energy in gasoline,
and it ended up blowing away in the wind.  You could have 
stayed home, and just opened some windows and let some
money blow away while you ate chips and watched TV.
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Answer:

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Explanation:

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the energy of the photons is not enough to carry out an electronic transition between two states of the material, when we decrease the wavelength (the energy of the photons increases), the point is reached where the energy of the beam is equal to some energy of a transition, by which the electrons are promoted and since we can see a certain charge, as the atoms are neutral, some electrons must be removed from the material, this is represented in the macroscopic case as the work function of the material, consequently a unbalanced load that is what we can measure.

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8 0
2 years ago
A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou
irakobra [83]

Answer:

λ = 3.2 x 10⁻⁷ m = 320 nm

Explanation:

The relationship between the velocity of electromagnetic waves (UV rays) and the their frequency is:

v = fλ

where,

v = c = speed of the electromagnetic waves (UV rays) = speed of light

c = 3 x 10⁸ m/s

f = frequency of the electromagnetic waves (UV rays) = 9.38 x 10¹⁴ Hz

λ = wavelength of the electromagnetic waves (UV rays) = ?

Therefore, substituting the values in the relation, we get:

3 x 10⁸ m/s = (9.38 x 10¹⁴ Hz)(λ)

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3 0
3 years ago
Read 2 more answers
An interference pattern is produced by light with a wavelength 580 nm from a distant source incident on two identical parallel s
irakobra [83]

Answer:

Explanation:

1 )

Here

wave length used that is λ = 580 nm

=580 x 10⁻⁹

distance between slit d = .46 mm

= .46 x 10⁻³

Angular position of first order interference maxima

= λ / d radian

= 580 x 10⁻⁹ / .46 x 10⁻³

= 0.126 x 10⁻² radian

2 )

Angular position of second order interference maxima

2 x  0.126 x 10⁻² radian

= 0.252 x 10⁻² radian

3 )

For intensity distribution the formula is

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For angular position of θ1

δ = .126 x 10⁻² radian

I = I₀ cos².126x 10⁻²/2

= I₀ X .998

For angular position of θ2

I = I₀ cos².126x2x 10⁻²/2

=  I₀ cos².126x 10⁻²

8 0
3 years ago
An electron is a particle with a ____.
Dahasolnce [82]

A beta particle. Hoped I help. Sorry if it wrong.

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