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irina [24]
3 years ago
14

What is an output force?

Physics
1 answer:
Rasek [7]3 years ago
4 0
An output for is the force a person applies to a simple machine.
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An electron travels west to east with a kinetic energy of 10 keV. The Earth's magnetic field in Pittsburgh is 19,911.5 nT in the
ch4aika [34]

Explanation:

It is given that,

Kinetic energy of the electron, E_k=10\ keV=10^4\ eV=1.6\times 10^{-15}\ J

Let the east direction is +x direction, north direction is +y direction and vertical direction is +z direction.    

The magnetic field in north direction, B_y=19911.5\ nT

The magnetic field in west direction, B_x=-3257.1\ nT

The magnetic field in vertical direction, B_z=48381.8 \ nT

Magnetic field, B=(-3257.1i+19911.5j+48381.8k)\ nT

Firstly calculating the velocity of the electron using the kinetic energy formulas as :

E_k=\dfrac{1}{2}mv^2

v=\sqrt{\dfrac{2E_k}{m}}

v=\sqrt{\dfrac{2\times 1.6\times 10^{-15}}{9.1\times 10^{-31}}}

v=5.92\times 10^7 i\ m/s (as it is moving from west to east)

The force acting on the charged particle in the magnetic field is given by :

F=q(v\times B)

F=1.6\times 10^{-19}\times (5.92\times 10^7 i\times (-3257.1i+19911.5j+48381.8k)\times 10^{-9})

Since, i\times j=k\ \\j\times k=i\\k\times i=j

And, i\times i=j\times j=k\times k=0

F=1.6\times 10^{-19}\times [1178 k-2864.20j]

|F|=1.6\times 10^{-19}\times \sqrt{1178^2+2864.20^2}

F=4.95\times 10^{-16}\ N

(b) Let a is the acceleration of the electron. It can be calculated as :

a=\dfrac{F}{m}

a=\dfrac{4.95\times 10^{-16}}{9.1\times 10^{-31}}

a=5.43\times 10^{14}\ m/s^2

Hence, this is the required solution.

4 0
3 years ago
Define investigation to show its scientific meaning.
Murrr4er [49]

Answer:

the action of investigating something or someone; formal or systematic examination or research.

Explanation:

This definition is provided by Oxford Languages

7 0
3 years ago
Convert 700kg into grams<br>​
Verdich [7]

Answer:

700,000g

Explanation:

1kg = 1000 Grams (or any kg x 1000)

5 0
2 years ago
Read 2 more answers
The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit
Margaret [11]

Let  us consider two bodies having masses m and m' respectively.

Let they are  separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as -  F = G\frac{mm'}{r^{2} }   where G is the gravitational force constant.

From the above we see that F ∝ mm' and F\alpha \frac{1}{r^{2} }

Let the orbital radius of planet  A is r_{1}  = r and mass of planet is m_{1}.

Let the mass of central star is m .

Hence the gravitational force for planet A  is f_{1} =G \frac{m_{1}*m }{r^{2} }

For planet B the orbital radius  r_{2} =2r_{1} and mass m_{2} = 3 m_{1}

Hence the gravitational force f_{2} =G\frac{m m_{2} }{r^{2} }

                                                 f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }

                                                 = \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }

Hence the ratio is  \frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2}  }{Gmm_{1}/r_{1} ^2 }

                                      =\frac{3}{4}     [ ans]


                                                 

                           

3 0
3 years ago
Read 2 more answers
Is 0.0 m/s squared increasing or decreasing
Agata [3.3K]
Decrease because its 0.0 m/s
4 0
3 years ago
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