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Grace [21]
4 years ago
8

Un autocar que circula a 81 km/h frena uniformemente con una aceleración de -4,5 m/s2.

Physics
1 answer:
horsena [70]4 years ago
8 0

Answer:

a) \Delta x=56.25 m

b) imagen adjunta

Explanation:

a) Primero debemos hacer la conversión de 81 km/h a m/s, esto es 22.5 m/s.

Ahora, usando la ecuacion cinemática, en un movimiento acelerado tenemos:

v_{f}^{2}=v_{0}^{2}+2a \Delta x

Queremos encontrar la posición hasta detenerse, osea vf = 0.

\Delta x=\frac{-v_{0}^{2}}{2a}

\Delta x=\frac{-22.5^{2}}{-2*4.5}

\Delta x=56.25 m

b) Para este caso el gráfico se encuentra adjunto.                                      

Espero que te sirva de ayuda!                                                                                                                                                                          

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g A rotating wheel requires 5.00 s to rotate 28.0 revolutions. Its angular velocity at the end of the 5.00-s interval is 96.0 ra
Setler [38]

Answer:

The angular acceleration of the wheel is 15.21 rad/s².

Explanation:

Given that,

Time = 5 sec

Final angular velocity = 96.0 rad/s

Angular displacement = 28.0 rev = 175.84 rad

Let \alpha be the angular acceleration

We need to calculate the angular acceleration

Using equation of motion

\theta=\omega_{i} t+\dfrac{1}{2}\alpha t^2

Put the value in the equation

175.84=\omega_{i}\times 5+\dfrac{1}{2}\times\alpha\times(5)^2

175.84=\omega_{i}\times 5+12.5\alpha......(I)

Again using equation of motion

\omega_{f}=\omega_{i}+\alpha t

Put the value in the equation

96.0=\omega_{i}+\alpha \times 5

On multiply by 5 in both sides

480=\omega_{i}\times 5+\alpha\times 25....(II)

On subtract equation (I) from equation (II)

480-175.84=\alpha(25-5)

304.16=\alpha\times20

\alpha=\dfrac{304.16}{20}

\alpha=15.21\ rad/s^2

Hence, The angular acceleration of the wheel is 15.21 rad/s².

5 0
3 years ago
Katie is ill with the flu. She immediately contacts her teacher to get the
Basile [38]

Answer: Planning

Explanation: cause i got it right

6 0
3 years ago
In your research lab, a very thin, flat piece of glass with refractive index 2.30 and uniform thickness covers the opening of a
trapecia [35]

Answer:

λ₀= 495.88 nm

Explanation:

To analyze this constructive interference interference experiment by reflection, let's look at two important aspects:

* when a ray of light passes from a medium with a lower index, they refact to another medium with a higher index, the reflected ray has a phase difference of pyres

* When a beam penetrates a material medium, the wavelength of the radiation changes according to the refractive index of the material.

       λₙ = λ₀ / n

when we introduce these aspects in the expression of contributory interference, it remains

        2 d sin θ = (m + ½) λ₀ / n

In general, reflection phenomena are measured at an almost normal angle, whereby θ = π/2  and sin θ = 1

        2 d = (m +1/2) λ₀/ n

         2n d = (m + ½) λ₀

Let's apply this expression to our case

         d = (m + ½) λ₀ / 2n

Suppose we measure on the first interference, this is m = 0

         d = ½ λ₀ / 2n

 

let's calculate

         d = ½ 496 10⁻⁹ / (2 2.30)

         d = 53.9 10-9 m  

This is the thickness of the glass, the next wavelength that gives constructive interference is

          λ₀ = 2 n d / (m + ½)

let's calculate

          λ₀ = 2 2.3 5.39 10-8 / (1 + ½)

          λ₀= 4.9588 10-7 m

          λ₀= 495.88 nm

5 0
4 years ago
How does noise affect signals? what happens if the level of noise becomes too high relative to the strength of the signal.
Nat2105 [25]

Noise could be defined as electromagnetic fields that affect analog signals that are constantly changing. This process does not occur in a similar way with digital signals, which have fixed electrostatic pulses (For this reason they are able to withstand 'noise' because the power of these signals are much stronger than the power coming from noise).

That phenomenon does not happen with the analog signals which have a variable intensity and become vulnerable to any electronic noise interference.

When very high electromagnetic fields are generated, the waves of the analog signal cannot be perceived which causes problems in the transmitted signal (making it unintelligible to the receiver)

6 0
4 years ago
If the light leaves a glass block ( n = 1.5 )with an angle of refraction of 70°. what is the angle of incidence?
julsineya [31]

Answer:

The angle of incidence is 38.8°.

Explanation:

let n1 = 1.5 be the refractive index of glass and ∅1 be the angle of incidence, let n2 be the  refractive index of air and ∅2 be the angle of refraction.

by Snell's law:

n1×sin(∅1) = n2×sin(∅2)

sin(∅1) = n2×sin(∅2)/n1

           = (1)×sin(70)/1.5

      ∅1 = 38.8°

Therefore, the angle of incidence is 38.8°.

5 0
3 years ago
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