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zvonat [6]
2 years ago
7

A person riding a bike has an initial velocity of 10 meters per second and accelerates at a constant rate of 2 meters per second

^2. After 5 seconds, the person would travel ___ meters.
Physics
1 answer:
Montano1993 [528]2 years ago
7 0

After 5 seconds, the person would have travel 75 m

<h3>What is acceleration? </h3>

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

  • a is the acceleration
  • v is the final velocity
  • u is the initial velocity
  • t is the time

<h3>How to determine the final velocity</h3>

We'll begin by obtaining the final velocity of the person. This can be obtained as follow:

  • Initial velocity (u) = 10 m/s
  • Acceleration (a) = 2 m/s²
  • Time (t) = 5 s
  • Final velocity (v) = ?

a = (v – u) / t

2 = (v – 10) / 5

Cross multiply

v – 10 = 2 × 5

v – 10 = 10

Collect like terms

v = 10 + 10

v = 20 m/s

<h3>How to determine the distance travelled</h3>

The distance travelled can be obtained as follow:

  • Initial velocity (u) = 10 m/s
  • Final velocity (v) = 20 m/s
  • Acceleration (a) = 2 m/s²
  • Distance (s) = ?

v² = u² + 2as

Collect like terms

2as = v² - u²

Divide both sides by 2a

s = (v² - u²) / 2a

s = (20² - 10²) / (2 × 2)

s = 75 m

Thus, the distance travelled by the person is 75 m

Learn more about velocity:

brainly.com/question/3411682

Learn more about acceleration:

brainly.com/question/491732

#SPJ1

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In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the
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Answer:

The wavelength is \lambda_2 =  534 *10^{-9} \ m

Explanation:

From the question we are told that

   The wavelength of the first light is  \lambda _ 1 =  587 \ nm

    The order of the first light that is being considered is  m_1  =  10

     The order of the second light that is being considered is  m_2  =  11

Generally the distance between the fringes for the first light is mathematically represented as

      y_1 =  \frac{ m_1  * \lambda_1 *  D}{d}

 Here  D is the distance from the screen

 and    d  is the distance of separation of the slit.

      For the second light the distance between the fringes is  mathematically represented as

         y_2 =  \frac{ m_2  * \lambda_2 *  D}{d}

Now given that both of the light are passed through the same double slit

       \frac{y_1}{y_2}  =  \frac{\frac{m_1 *  \lambda_1 * D}{d}  }{\frac{m_2 *  \lambda_2 * D}{d}  } = 1

=>    \frac{ m_1 *  \lambda _1  }{ m_2  *  \lambda_2} =  1

=>     \lambda_2 =  \frac{m_1 *  \lambda_1}{m_2}

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Answer:

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Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

T = 2\pi \sqrt{\frac{r^3}{GM}}

now for the time period of moon around the earth we can say

T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}

here we know that

T_1 = 0.08 year

r_1 = 0.0027 AU

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Now if the same formula is used for revolution of Earth around the sun

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here we know that

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M_s = mass of Sun

now we have

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