All categories

3 years ago

What is the the steadiest firing position?

Physics

1 answer:

Paha777 [63]3 years ago

3 0

The closer you are to the ground the more accurate you'll be. That's why most snipers are in the "prone" position.

You might be interested in

The blood pressure in the human body is greater at the feet than at the brain

GarryVolchara [31]

Blood pressure is greater in feet because of gravity

5 0

3 years ago

Read 2 more answers

The what’s the difference does it mean velocity and vector Quantity is the same

ziro4ka [17]

Answer:

No they are totally different...

Velocity is displacement/time. while vector is both magnitude and direction.....

Velocity is a vector quantity because it has both magnitude and direction

6 0

3 years ago

A particular HeNe laser beam at 633 nm has a spot size of 0.8 mm. Assuming a

Kobotan [32]

Answer:

Explanation:

8 0

3 years ago

Please help meeeee!!!!!!

Vinil7 [7]

Answer:

Net force

Explanation:

Bruh, easy question

3 0

2 years ago

What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wa

dlinn [17]

Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

(a). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}$

$E=3.315\times10^{-17}\ J$

(b). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}$

$E=6.709\times10^{-21}\ J$

(c). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}$

$E=3.315\times10^{-11}\ J$

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}$

$E=6.709\times10^{-9}\ J$

Hence, This is the required solution.

6 0

3 years ago