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klemol [59]
2 years ago
10

It takes a minimum distance of 76.50 m to stop a car moving at 15.0 m/s by applying the brakes (without locking the wheels). Ass

ume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 32.0 m/s.
Physics
1 answer:
Vinvika [58]2 years ago
5 0

The minimum stopping distance when the car is moving at 32.0 m/s is 348.3 m.

<h3>Acceleration of the car </h3>

The acceleration of the car before stopping at the given distance is calculated as follows;

v² = u² + 2as

when the car stops, v = 0

0 = u² + 2as

0 = 15² + 2(76.5)a

0 = 225 + 153a

-a = 225/153

a = - 1.47 m/s²

<h3>Distance traveled when the speed is 32 m/s</h3>

If the same force is applied, then acceleration is constant.

v² = u² + 2as

0 = 32² + 2(-1.47)s

2.94s = 1024

s = 348.3 m

Learn more about distance here: brainly.com/question/4931057

#SPJ1

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n ultraviolet light beam having a wavelength of 130 nm is incident on a molybdenum surface with a work function of 4.2 eV. How f
pashok25 [27]

Answer:

The speed of the electron is 1.371 x 10⁶ m/s.

Explanation:

Given;

wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m

the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J

The energy of the incident light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f = c / λ

E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{130*10^{-9}} \\\\E = 15.291*10^{-19} \ J

Photo electric effect equation is given by;

E = W₀ + K.E

Where;

K.E is the kinetic energy of the emitted electron

K.E = E - W₀

K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J

K.E = 8.563 x 10⁻¹⁹ J

Kinetic energy of the emitted electron is given by;

K.E = ¹/₂mv²

where;

m is mass of the electron = 9.11 x 10⁻³¹ kg

v is the speed of the electron

v = \sqrt{\frac{2K.E}{m} } \\\\v =  \sqrt{\frac{2*8.563*10^{-19}}{9.11*10^{-31}}}\\\\v = 1.371 *10^{6} \ m/s

Therefore, the speed of the electron is 1.371 x 10⁶ m/s.

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4 years ago
PLEASE HELP ME ANSWER DUE SOON BRAINLIEST FOR WHOEVER
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Light from the star Betelgeuse takes 640 years to reach Earth. How far away is Betelgeuse in units of light-years? Name any hist
Eduardwww [97]

Answer:

Betelgeuse is 640 light years away from earth.

Explanation:

A light-year is an astronomical unit to measure the distance the light travels in a calendar year.

If the light from a star takes 640 years to reach us, then that star its 640ly away from us.

Betelgeuse has been labeled as a Variant Star, which means that its brightness can fluctuate over the course of years, this has made difficult for astronomers to measure the exact distance of the star. Right now the star is estimated to be around 613 and 881ly away from earth, although, for the sake of your second question, we will take 640 years as our estimated value.

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6 0
3 years ago
Read 2 more answers
Answer ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀​
ikadub [295]

\sf \huge \purple{ Question : -  }

Initial velocity of a car is 36 km/h . Find the distance after min, if it goes with acceleration 2 m/s².

\sf \huge \color {gold}{ AnSwer :- }

Initial velocity, u = 36 km/h

\sf u = 36 \times  \dfrac{5}{18}  {ms}^{ - 1}  = 10 {ms}^{ - 1}

Time, t = 1min

  • t = 60s

Acceleration, a = 2m/s²

Apply 2nd equation of motion

\longrightarrow \sf s = ut +  \dfrac{1}{2} at ^{2}  \\  \\  \longrightarrow \sf s =(10)(60) +  \frac{1}{2} (2)(60) {}^{2}  \\  \\  \longrightarrow \sf s =600 + 3600   \\  \\  \longrightarrow \sf  \blue{ s =4200m}  \:  \: \large \blue \star

6 0
3 years ago
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