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Kitty [74]
3 years ago
13

Calculate the height from from which a body is released from rest if its velocity just before hitting the ground is30m\s

Physics
1 answer:
Kamila [148]3 years ago
5 0

Answer:

height = 45 m

Explanation:

Initial velocity = 0 m/s

Final velocity = 30 m/s

Acceleration due to gravity = +10m/s^2 ( because ball is going down )

Using the eqn of motion

{v}^{2}  -  {u}^{2}  = 2gh

where v= final velocity ; u = initial velocity ; g = acceleration due to gravity ; h = height

So,

{30}^{2}  -  {0}^{2}  = 2 \times 10 \times h

=  > 20h = 900

=  > h =  \frac{900}{20}  = 45

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An objects weighs 30n on earth a second object weighs 30n on thee moon which has the greatest mass of is it the same
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Which option correctly matches the chemical formula of a compound with its name?
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In a simplified model of a hydrogen atom, the electron moves around the proton nucleus in a circular orbit of radius 0.53×10−10m
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Answer

given,

radius of the circular orbit, r = 0.53 x 10⁻¹⁰ m

mass of electron, M = 9.11 x 10⁻³¹ Kg

charge of electron, q₁ = 1.6 x 10⁻¹⁹ C

                                q₂ = 1.6 x 10⁻¹⁹ C

we know, force between two charges

F = \dfrac{kq_1q_2}{r^2}

F = \dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.53\times 10^{-10})^2}

  F = 8.20 x 10⁻⁸ N

b) using newton's second law

F = m a

m a =  8.20 x 10⁻⁸

a =\dfrac{8.20\times 10^{-8}}{9.11\times 10^{-31}}

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 a =\dfrac{v^2}{r}

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d) the period of the circular motion.

    T=\dfrac{2\pi}{\omega}

    T=\dfrac{2\pi r}{v}

    T=\dfrac{2\pi\times 0.53\times 10^{-10}}{2.18\times 10^6}

          T = 1.53 x 10⁻¹⁶ s

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