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katen-ka-za [31]
2 years ago
8

What is the formula of acceleration?

Physics
2 answers:
kobusy [5.1K]2 years ago
5 0

Answer:

\overline{a} = \frac{v - v_0}{t} = \frac{\Delta v}{\Delta t}

\overline{a} = average acceleration

v = final velocity

v_0 = starting velocity

t = elapsed time

Explanation:

Please give me brainliest :)

kherson [118]2 years ago
3 0

Answer:

\overline{a} = \frac{v - v_0}{t} = \frac{\Delta v}{\Delta t}

\overline{a} = average acceleration

v = final velocity

v_0 = starting velocity

t = elapsed time

Explanation:

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What are the different types of Energy transformations and Energy Transfers?
Harman [31]

Explanation:

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The Sun transforms nuclear energy into heat and light energy.

Our bodies convert chemical energy in our food into mechanical energy for us to move.

An electric fan transforms electrical energy into kinetic energy.

7 0
3 years ago
The acceleration due to gravity on the moon is about 5.4ft/s2. if your weight is 150lbf on earth
Ivenika [448]

... then your weight is <em>25.2 lbf</em> on the moon.

6 0
3 years ago
Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consid
Korolek [52]

Answer:

a

    z= 2.5 \ m

b

   z =  (1 \ m ,  4 \ m )

Explanation:

From the question we are told that

     Their distance apart is  d =  5.00 \ m

      The  wavelength of each source wave \lambda =  6.0 \ m

Let the distance from source A  where the construct interference occurred be z

Generally the path difference for constructive interference is

              z - (d-z) =  m \lambda

Now given that we are considering just the straight line (i.e  points along the line connecting the two sources ) then the order of the maxima m =  0

  so

        z - (5-z) =  0

=>     2 z - 5 =  0

=>     z= 2.5 \ m

Generally the path difference for destructive  interference is

           |z-(d-z)| = (2m + 1)\frac{\lambda}{2}

=>         |2z - d |= (0 + 1)\frac{\lambda}{2}

=>        |2z - d| =\frac{\lambda}{2}

substituting values

          |2z - 5| =\frac{6}{2}

=>      z  =  \frac{5 \pm 3}{2}

So  

      z =  \frac{5 + 3}{2}

      z =  4\ m

and

      z =  \frac{ 5 -3 }{2}

=>   z =  1 \ m

=>    z =  (1 \ m ,  4 \ m )

7 0
3 years ago
lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
qaws [65]

2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

5 0
2 years ago
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