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Aliun [14]
3 years ago
14

And this one as well

Physics
1 answer:
Serggg [28]3 years ago
5 0
If I'm correct both of the answers are A.
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3 a A motorcyclist starts from rest and reaches
andrew-mc [135]

The acceleration of the body is 2 m/s^2 while the deceleration is - 1.2 m/s^2.

<h3>What is the acceleration?</h3>

Let us recall that the acceleration is the change in the speed of a body with time. We have been told that the body accelerates for 3s and then decelerates to 2s. This implies that the total time that the object spent in motion is 5 s.

Thus;

v = u + at

v = final velocity

u = initial velocity

a = acceleration

t = time taken

v - u/t = a

a = 6 - 0/3

= 2 m/s^2

Again;

v - u/t = a

a = 0 - 6/5

a = - 1.2m/s^2

Learn more about acceleration:brainly.com/question/12550364

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<h3 />
7 0
2 years ago
Calculate the wavelength of a photon of electromagnetic radiation with a frequency of 151.7 mhz
AlekseyPX

The wavelength of a photon of electromagnetic radiation with a frequency of 151.7 Mhz is 1.978 meter

According to the question

The frequency of a photon of electromagnetic radiation = 151.7 Mhz

By using the Formula of wavelength and frequency

λ = C/f

Where,

λ (Lambda) = Wavelength in meters

C = Speed of Light (c = 3x10⁸ m/s)

f = Frequency

Now, the wavelength of a photon of electromagnetic radiation with a frequency of 151.7 Mhz is  

1 Mhz = 1000000 Hz

151.7 Mhz = 151.7 * 10⁶ Hz

λ = C/f  

Substituting the value of C and f  

  λ =  \frac{3*10^{8} }{151.7 * 10^{6}  }

  λ = 1.978 meter

Hence,  the wavelength of a photon of electromagnetic radiation with a frequency of 151.7 Mhz is 1.978 meter

To know more about  wavelength and frequency here:

brainly.com/question/18651058

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5 0
2 years ago
An electron in an atom's orbital shell, labeled X in the model below, released enough energy to move to a different orbital shel
Delicious77 [7]

Answer:

Lower energy shell which will be nearer to the nucleus.

Explanation:

When electron move from one energy level to another, an electron must gain or lose just the right amount of energy.

When atoms releases energy, electrons move into lower energy levels.  The electrons in the shells aways from the nucleus have more energy as compared to the electrons in the nearer shells.

Electrons with the lowest energy are found closest to the nucleus, where the attractive force of the positively charged nucleus is the greatest. Electrons that have higher energy are found further away

7 0
3 years ago
Which of these is not a conservative force and does not have a form of potential energy connected with it?
a_sh-v [17]
Friction I believe is the answer

hope this helps
6 0
4 years ago
a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate &lt;br /&
Katen [24]

Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

        a = v² / r

we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

a) we substitute in 1

         T = \frac{mg }{cos 30}

         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

8 0
3 years ago
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