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anyanavicka [17]
2 years ago
5

A car with a velocity of 22 m/s is accelerated at a rate of 1.6m/s2 for 6.8s. determine the final velocity

Physics
1 answer:
Ivahew [28]2 years ago
3 0

A car with a velocity of 22 m/s is accelerated at a rate of 1.6 m/s^2 for 6.8s has the final velocity t be 32.88 m/s.

The acceleration means the amount of velocity changing per unit time.

The given data:

initial velocity, u = 22 m/s

time, t = 6.8 s

acceleration, a = 1.6 m/s^2

We will be using the equation of motion:

v = u + at

\therefore v=22+1.(6.8)

\Rightarrow v=22+10.88

\Rightarrow v=32.88 \ m/s

The final velocity become 32.88 m/s.

To learn more about Attention here:

https://brainly.in/question/10557838

#SPJ4

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Compare the gravitational acceleration on the following objects compared to the Sun using:
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The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765  x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}

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3 0
3 years ago
Light of wavelength 520 nm is used to illuminate normally two glass plates 21.1 cm in length that touch at one end and are separ
Oksi-84 [34.3K]

Answer:

The number is  Z = 216 \ fringes

Explanation:

From the question we are told that

      The wavelength is  \lambda  =  520 \ nm =  520 *10^{-9} \ m

       The length of the glass plates is y  = 21.1cm = 0.211 \ m

      The distance between the plates (radius of wire ) =  d =  0.028 mm  =  2.8 *10^{-5} \  m

   Generally the condition for constructive  interference in a film is mathematically represented as

            2 *  t  = [m +  \frac{1}{2}  ]\lambda

Where  t is the thickness of the separation between the glass i.e  

    t  = 0 at the edge where the glasses are touching each other and  

     t =  2d at the edge where the glasses are separated by the wire  

   m is the order of the fringe it starts from  0, 1 , 2 ...

So  

       2 *  2 * d   = [m +  \frac{1}{2}  ] 520 *10^{-9}

=>   2 *  2 *   (2.8 *10^{-5}) = [m +  \frac{1}{2}  ] 520 *10^{-9}

=>    

       m = 215

given that we start counting m from zero

   it means that the number of  bright fringes that would appear is

         Z =  m + 1

=>    Z =  215 +1

=>     Z = 216 \ fringes

3 0
3 years ago
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