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Viefleur [7K]
2 years ago
11

The velocity of particle varies with time as equation

Physics
2 answers:
hoa [83]2 years ago
6 0

Answer:

Don't know if you meant to write 3t²i - 2tj

If so

a=dv/dt = 6ti - 2j

at t=1sec

a = 6(1)I - 2j

a=6i - 2j

Magnitude of a = √6²+(-2)²

a= 2√10ms-²

or 6.32ms-².

Advocard [28]2 years ago
3 0

Velocity of a particle varies with its displacement as v = ( √(9 ... Velocity of a particle varies with its displacement as v = ( √(9 - x^2) ) m/sFind the magnitude of maximum acceleration of the particle.

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Explanation:

Nowadays we know <u>cells are essential microscopic units that make up the living beings, capable of reproducing independently. </u>

However, this is the result of a long process of discoveries and studies made since the 19th century, in which the continuous improvement of new technologies was helpful.

In fact, it is wel known the English scientist Robert Hooke was the first to discover the existence of cells by looking through a compound microscope at a cork sheet, realizing that it was made up of small polygonal holes (like those of a honeycomb) that reminded him of the chambers in which the monks stayed (called cells). Then, during the next centuries more studies were made until we had the current knowledge about the structure of a cell.

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2 years ago
Where is the frequency of ultrasound in relation to the range of human ability to hear
kykrilka [37]

Answer:

ultra sounds have frequency higher than the upper audible limit of human hearing, for healthy, young adults.

Explanation:

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2 years ago
An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p
igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

Explanation:

Given:

Distance from the wire to the field point r = 2.83 \times 10^{-2} m

Speed of electron v = 35.5 \%c

Current I = 17.7 A

For finding the acceleration,

First find the magnetic field due to wire,

  B = \frac{\mu _{o}I }{2\pi r }

Where \mu_{o} = 4\pi   \times 10^{-7}

  B = \frac{4\pi \times 10^{-7}  \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }

  B = 12.50 \times 10^{-5} T

The magnetic force exerted on the electron passing through straight wire,

  F = qvB  

  F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}

  F = 21.3 \times 10^{-16} N

From the newton's second law

  F = ma

Where m = mass of electron = 9.1 \times 10^{-31} kg

So acceleration is given by,

   a = \frac{F}{m}

   a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }

   a = 2.34 \times 10^{15} \frac{m}{s^{2} }

Therefore, the magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

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Which substance is uniquely a product of incomplete combustion of a fossil fuel?
Serga [27]

I can think of two of them:

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