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katrin2010 [14]
2 years ago
12

Think of a famous person and gather information about him/her​

Physics
1 answer:
dybincka [34]2 years ago
4 0

Abraham Lincoln

Explanation:

hope it helps you a little

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A 3.2 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 4 + 15.7 x − 1.5 x 2 , where Fx
garik1379 [7]

Answer:

Explanation:

Work: This can be defined as the product of force and distance. The unit of work is Joules (J). it can be expressed mathematically as

W = F×d

or

W = \int\limits^b_a {Fx} \, dx.................................. Equation 1

Where b = upper limit, a = lower limit, Fx = expression of force.

<em>Given: a = 0 , b = 1.3 m, Fx = 4 + 15.7x - 1.5x²</em>

Substituting these values into equation 1

<em>W = \int\limits^a_b {(4 + 15.7x - 1.5x^{2} )dx} \,</em>

W = ᵇ[4x + 15.7x²/2-1.5x³/3 +C]ₐ

Work = upper limit - lower limit

Work = ᵃ[4x + 15.7x²/2 - 1.5x³/3 +C] - [4x + 15.7x²/2 + 1.5x³/3 +C]ᵇ............... Equation 2

Substituting the values of a and b into equation 2

Work = [4(1.3) + 15.7(1.3)²/2-1.5(1.3)³/3 + C] - [0 +C]

Work = [5.2 + 26.53 -3.29 + C] - C

Work = 28.44 J

Work done by the force = 28.44 J.

8 0
3 years ago
Dropping a stone from a height is work or not?​
kvasek [131]

Answer:

Dropping a stone from a height is work  

Explanation:

Work is the action of a force through a distance.

w = fd

The Earth's gravitational attractive force is pulling the stone to its centre, so the Earth is doing work on the stone.

7 0
3 years ago
An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea
melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, v=1.37\times10^7\ \rm m/s

Explanation:

<u>Given:</u>

  • Linear charge density of the ring=0.1\ \rm \mu C/m
  • Radius of the ring R=0.2 m
  • Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C

Potential due the ring at a distance x from the centre of the rings is given by

V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V

Let\Delta U be the change in potential Energy given by

\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s

So the electron will be moving with v=1.37\times10^7\ \rm m/s

5 0
3 years ago
Convert 5g/cm^3 into kg/m^3​
Sliva [168]

Answer:

0.01135624

Explanation:

7 0
2 years ago
Which of the following codes is used to report the removal of 37 skin tags by electrosurgical destruction?
Tema [17]
Are there any answer choices??
7 0
3 years ago
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