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Marina86 [1]
2 years ago
8

A flea jumps straight up to a maximum height of 0.400 m . what is its initial velocity v0 as it leaves the ground?

Physics
1 answer:
timama [110]2 years ago
8 0

For an object`s motion, the Kinematic equation is,

v^2=v_{0}^2+2ah

Here, v is the final velocity and h is stands for the height of the object and a is the acceleration of the object.

As according to question,

v=0m/s,a=g-9.8 m/s^2 and h = 0.400 m

Thus, putting these values in above equation, we get

0= v_{0}^2 -2gh

or

v_{0} =\sqrt{2 \times 9.8 \times 0.400 }

v_{0} = 2.8 m/s

Therefore, initial velocity is 2.8 m/s



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liubo4ka [24]

The collision is a form of inelastic collision because the it forms a single mass after is collides. So it can be solve by momentum balance

( 0.08 kg * 50 m/s ) + ( 0.06 kg * 50 m/s) = ( 0.08 + 0.06 kg ) v

V = 50 m/s

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2 years ago
A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac
GREYUIT [131]

Answer:

The flux through the surface of the cube is 2.314\ Nm^{2}/C

Solution:

As per the question:

Edge of the cube, a = 8.0 cm = 8.0\times 10^{- 2}\ m

Volume Charge density, \rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}

Now,

To calculate the electric flux:

\phi = \frac{q}{\epsilon_{o}}                                                      (1)

where

\phi = electric flux

\epsilon_{o} = 8.85\times 10^{- 12}\ F/m = permittivity of free space  

Volume Charge density for the given case is given by the formula:

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Volume of cube, V = a^{3}

Thus

V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}

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q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}

q = 2.048\times 10^{-11}\ F = 20.48\ pF

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Answer:

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Explanation:

Data provided;

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= 59.9 × 10³ V

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on substituting the respective values, we get

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