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Kitty [74]
3 years ago
5

When a force is applied to a wheel, its axle exerts a greater force?

Physics
1 answer:
makvit [3.9K]3 years ago
7 0

Answer:

That is true.

Explanation:

When the input force is applied to the wheel, as it is with a doorknob, the axle turns over a shorter distance but with greater force, so the mechanical advantage is greater than 1.

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the net force on a vehicle is accelerating at a rate of 1.5 milliseconds is 1800 Newtons. What is the mass of the vehicles neare
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Answer:

The mass is 1200 kilograms

Explanation:

Because Force is equal to mass times acceleration (F=m×a)

F=m×a

1800N=?×1.5

1800÷1.5=1200

1800N=1200Kg×1.5

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Rank the wavelengths of the following quantum particles from the largest to the smallest. If any have equal wavelengths, display
goldfiish [28.3K]

The wavelengths of the following quantum particles from the largest to the smallest is (d) > (a) = (e) > (b) > (c)

  • De Broglie proposed that because light has both wave and particle properties, matter exhibits both wave and particle properties. This property has been explained as the dual behavior of matter.
  • From his observations, de Broglie derived the relationship between the wavelength and momentum of matter. This relationship is known as de Broglie's relationship.

De Broglie's relationship is given by  \lambda=\frac{h}{mv} .

This can be written as \lambda=\frac{h}{p}  .....(1) where λ  is known as de Broglie wavelength and p is momentum , h = Plank’s constant .

As we know that mass of proton is greater than electron and photon .

(a)For photon , the momentum is given by p=\frac{E}{c}     ...(2)  where c is the speed is the speed of light .

Putting E = 3eV in equation (2) , we get

              p=\frac{3\times 1.6\times10^-^1^9}{3\times 10^8} \\p=1.6\times10^-^2^7Js/m

Putting this value of p in equation (1) , we get

  \lambda=\frac{6.62\times10^{-34}}{1.6\times10^{-27}}\\\lambda=4.13\times10^{-7}

(b) As we know that  kinetic energy is given by

         K.E=\frac{1}{2} mv^2\\\\2K.E = mv^2\\2K.E\times m = m^2v^2\\2K.E\times m = (mv)^2\\2K.E\times m = p^2\\\\\sqrt{2K.E\times m }=p      ...(3)

Where mass of electron is 9.1\times10^-^3^1 kg .

Putting K.E = 3eV in equation (3) , we get

    \sqrt{2K.E\times m }=p\\p=\sqrt{2\times3\times1.6\times10^{-19} \times 9.31\times10^{-31} }\\p=\sqrt{89.376\times10^{-50}} \\p=9.45\times10^{-25}Js/m

Putting this value of p in equation (1) , we get

\lambda=\frac{6.62\times10^{-34}}{9.45\times10^{-25}}\\\lambda=0.7005\times10^{-9}

(c) Putting m=1.67\times10^{-27}kg and K.E = 3eV in equation (3) , we get

     \sqrt{2K.E\times m }=p\\p=\sqrt{2\times3\times1.6\times10^{-19} \times 1.67\times10^{-27} }\\p=\sqrt{16.032\times10^{-46}} \\p=4.003\times10^{-23}Js/m

Putting this value of p in equation (1) , we get

\lambda=\frac{6.62\times10^{-34}}{4.003\times10^{-23}}\\\lambda=1.65\times10^{-11}

(d) Putting E = 0.3eV in equation (2) , we get

   p=\frac{0.3\times 1.6\times10^-^1^9}{3\times 10^8} \\p=1.6\times10^-^2^8Js/m

Putting this value of p in equation (1) , we get

\lambda=\frac{6.62\times10^{-34}}{1.6\times10^{-28}}\\\lambda=4.13\times10^{-6}

(e) Putting p=3eV/c in equation (1) , we get

    \lambda=\frac{6.62\times10^{-34}\\\times3\times10^8}{3\times1.6\times10^{-19}}\\\lambda=4.13\times10^{-7}

On comparing the wavelength order should be (d) > (a) = (e) > (b) > (c) .

Learn more about de brogile here :

brainly.com/question/28165547

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Systematic is just a certain way of doing something. So science being systematic may just refer to the scientific method or how theories are proved in science by repeating the scientific method over and over again.
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4 years ago
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