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Arlecino [84]
2 years ago
9

A(n) 15.9 kg rock is on the edge of a(n

Physics
1 answer:
Mariulka [41]2 years ago
5 0

The potential energy (P.E) of the rock relative to the base of the cliff is 31,054.926 Joules..

<u>Given the following data:</u>

  • Mass of rock = 7.5 kg
  • Height = 0.7 m
  • Acceleration due to gravity = 9.8 m/s^2

To calculate the potential energy (P.E) of the rock relative to the base of the cliff:

<h3>What is potential energy?</h3>

Potential energy (P.E) can be defined as the energy that is possessed by an object due to its position (height) above planet Earth.

Mathematically, potential energy (P.E) is given by this formula;

P.E = mgh

<u>Where:</u>

  • P.E is the potential energy.
  • m is the mass of an object.
  • g is the acceleration due to gravity.
  • h is the height of an object.

Substituting the given parameters into the formula, we have;

P.E = 15.9 \times 9.8 \times 199.3

P.E = 31,054.926 Joules.

Read more on potential energy here: brainly.com/question/8664733

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Answer:

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Explanation:

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Adult men have an average height of 69.0 inches with standard deviation of 2.8 inches fins the night of a man with a z-score of
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Your low-flow showerhead is delivering water at 1.2×10−4m3/s, about 2.0 gallons per minute.
Oksanka [162]

To solve this problem it is necessary to apply the fluid mechanics equations related to continuity, for which the proportion of the input flow is equal to the output flow, in other words:

Q_1 = Q_2

We know that the flow rate is equivalent to the velocity of the fluid in its area, that is,

Q = VA

Where

V = Velocity

A = Cross-sectional Area

Our values are given as

Q_2 = 1.2*10^{-4}m^3/s

d = 0.021m

r = \frac{0.021}{2} = 0.0105m

Since there is continuity we have now that,

V_1A_1 = Q_2

V_1A_1 = 1.2*10^{-4}

V_1 = \frac{1.2*10^{-4}}{A_2}

V_1 = \frac{1.2*10^{-4}}{\pi r^2}

V_1 = \frac{1.2*10^{-4}}{\pi (0.0105)^2}

V_1 =0.347m/s

Therefore the speed of the water's house supply line is 0.347m/s

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3 years ago
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
Ksenya-84 [330]

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}}

    a = 9 x 10²² m/s²

c) speed of the electron

 a =\dfrac{v^2}{r}

 9\times 10^{22} =\dfrac{v^2}{0.53\times 10^{-10}}

   v² = 4.77 x 10¹²

  v = 2.18 x 10⁶ m/s

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

8 0
2 years ago
Read 2 more answers
The density of molybdenum is 10.28 g/cm^3 and it crystallizes in the face centered cubic unit cell. Calculate the edge length of
Effectus [21]

The edge length of the unit cell at the given atomic mass and density of the molybdenum is 314.2 pm.

<h3>Volume of molybdenum</h3>

V = (zm/ρN)

where;

  • z is 2 for cubic unit cell
  • m is mass of the molybdenum
  • ρ is density of the molybdenum

V = (2 x 95.96) / (10.28 x 6.02 x 10²³)

V = 3.10 x 10⁻²³ cm³

<h3>Edge length of the unit cell</h3>

a³ = V

a = (V)^¹/₃

a = ( 3.10 x 10⁻²³)^¹/₃

a = 3.142 x 10⁻⁸ cm

a = 3.142 x 10⁻¹⁰ m

a = 314.2 x 10⁻¹² m

a = 314.2 pm

Thus, the edge length of the unit cell at the given atomic mass and density of the molybdenum is 314.2 pm.

Learn more about edge length here:

brainly.com/question/16673486

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1 year ago
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