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yan [13]
3 years ago
9

4) A racing car undergoing constant acceleration covers 140 m in 3.6 s. (a) If it’s moving at 53 m/s at the end of this interval

, what was its speed at the beginning of the interval? (b) How far did it travel from rest to the end of the 140-m distance
Physics
1 answer:
Scorpion4ik [409]3 years ago
3 0

Answer

given,

distance = 140 m

time, t = 3.6 s

moving speed = 53 m/s

a) distance = (average velocity) x time

    D = \dfrac{v_0 + v_1}{2}\times t

    140 = \dfrac{v_0 + 53}{2}\times 3.6

       v₀ + 53 = 77.78

        v₀ = 24.78 m/s or 25 m/s

b) a = \dfrac{v-u}{t}

   a = \dfrac{53-25}{3.6}

         a = 7.8 m/s²

using equation of motion

  v₀² = v₁² + 2 a s

  53² = 0²+ 2 x 7.8 x s

  s = 180 m

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arlik [135]

Answer:

El neumático soportará una presión de 1.7 atm.

Explanation:

Podemos encontrar la presión final del neumático usando la ecuación del gas ideal:

PV = nRT

En donde:

P: es la presión

V: es el volumen

n: es el número de moles del gas

R: es la constante de gases ideales

T: es la temperatura

Cuando el neumático soporta la presión inicial tenemos:

P₁ = 1.5 atm

T₁ = 300 K

V_{1} = \frac{nRT_{1}}{P_{1}}  (1)  

La presión cuando T = 67 °C es:

P_{2} = \frac{nRT_{2}}{V_{2}}   (2)

Dado que V₁ = V₂  (el volumen del neumático no cambia), al introducir la ecuación (1) en la ecuación (2) podemos encontrar la presión final:

P_{2} = \frac{nRT_{2}}{V_{2}} = \frac{nRT_{2}}{\frac{nRT_{1}}{P_{1}}} = \frac{P_{1}T_{2}}{T_{1}} = \frac{1.5 atm*(67 + 273)K}{300 K} = 1.7 atm  

Por lo tanto, si en el transcurso de un viaje las ruedas alcanzan una temperatura de 67 ºC, el neumático soportará una presión de 1.7 atm.

Espero que te sea de utilidad!

4 0
3 years ago
A solar eclipse will occur Group of answer choices
myrzilka [38]

Answer:

3. at new Moon only when the Moon is on the ecliptic.

Explanation:

  • Solar eclipse is the condition when the moon comes in between the sun and the earth. In this condition the moon casts its shadow on the earth.
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  • As we know that the orbit of the earth around the sun and the orbit of the moon around the earth is elliptical which leads to a variation in the distance from their rotating centers, so not of every eclipse the moon covers the sun completely developing an annular eclipse.
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7 0
3 years ago
A student who weighs 750 newton’s runs up the steps (which have a height of 8 meters) in 13.5 seconds. How much work did the stu
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2 years ago
Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to
jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x 10^{-8} J.

(b) The wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency of the photon is 2.47 x 10^{25} Hz.

Explanation:

Let;

E_{1} = -13.60 ev

E_{2} = -3.40 ev

(a) Energy of the emitted photon can be determined as;

E_{2} - E_{1} = -3.40 - (-13.60)

           = -3.40 + 13.60

           = 10.20 eV

           = 10.20(1.6 x 10^{-9})

E_{2} - E_{1} = 1.632 x 10^{-8} Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x 10^{-8} Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x 10^{-34} Js), c is the speed of light (3 x 10^{8} m/s) and λ is the wavelength.

10.20(1.6 x 10^{-9}) = (6.6 x 10^{-34} * 3 x 10^{8})/ λ

λ = \frac{1.98*10^{-25} }{1.632*10^{-8} }

  = 1.213 x 10^{-17}

Wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x 10^{-8}  = 6.6 x 10^{-34} x f

f = \frac{1.632*10^{-8} }{6.6*10^{-34} }

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garik1379 [7]

85 protons are in astatine have.

5 0
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