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otez555 [7]
3 years ago
10

If 31.25

Physics
2 answers:
dlinn [17]3 years ago
5 0

Answer:

it will become charged by Q = 5C

Explanation:

When electron is removed from matter then the system will get positively charged

So we will have

Q = Ne

here N = number of electrons that is removed

N = 31.25 \times 10^{18}

also we know that charge on one electron is given as

e = 1.6 \times 10^{-19}

here we have

Q = Ne

Q = (31.25 \times 10^{18})(1.6 \times 10^{-19})

Q = 5 C

so it will become charged by Q = 5C

valentina_108 [34]3 years ago
4 0
<span>1 C = 6.24150965(16)×10^18 electrons

31.25 x 10^18 electrons / (6.24150965(16)×10^18 electrons / C) = 5.007 Coulombs

</span><span>I hope this helps. </span>
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The photon energies used in different types of medical x-ray imaging vary widely, depending upon the application. Single dental
pav-90 [236]

A) 5.0\cdot 10^{-11} m

The energy of an x-ray photon used for single dental x-rays is

E=25 keV = 25,000 eV \cdot (1.6\cdot 10^{-19} J/eV)=4\cdot 10^{-15} J

The energy of a photon is related to its wavelength by the equation

E=\frac{hc}{\lambda}

where

h=6.63\cdot 10^{-34}Js is the Planck constant

c=3\cdot 10^8 m/s is the speed of light

\lambda is the wavelength

Re-arranging the equation for the wavelength, we find

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4\cdot 10^{-15}J}=5.0\cdot 10^{-11} m

B) 2.0\cdot 10^{-11} m

The energy of an x-ray photon used in microtomography is 2.5 times greater than the energy of the photon used in part A), so its energy is

E=2.5 \cdot (4\cdot 10^{-15}J)=1\cdot 10^{-14} J

And so, by using the same formula we used in part A), we can calculate the corresponding wavelength:

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1\cdot 10^{-14}J}=2.0\cdot 10^{-11} m

4 0
2 years ago
A shell is fired with a horizontal velocity in the positive x direction from the top of an 80 m high cliff. The shell strikes th
Sonbull [250]

Answer:

V = 331.59m/s

Explanation:

First we need to calculate the time taken for the shell fire to hit the ground using the equation of motion.

S = ut + 1/2at²

Given height of the cliff S = 80m

initial velocity u = 0m/s²

a = g = 9.81m/s²

Substitute

80 = 0+1/2(9.81)t²

80 = 4.905t²

t² = 80/4.905

t² = 16.31

t = √16.31

t = 4.04s

Next is to get the vertical velocity

Vy = u + gt

Vy = 0+(9.81)(4.04)

Vy = 39.6324

Also calculate the horizontal velocity

Vx = 1330/4.04

Vx = 329.21m/s

Find the magnitude of the velocity to calculate speed of the shell as it hits the ground.

V² = Vx²+Vy²

V² = 329.21²+39.63²

V² = 329.21²+39.63²

V² = 108,379.2241+1,570.5369

V² = 109,949.761

V = √ 109,949.761

V = 331.59m/s

Hence the speed of the shell as it hits the ground is 331.59m/s

7 0
3 years ago
A child is perched perilously above a 150m ravine. The child has a mass of 45kg. How much
Finger [1]

Answer:

<em>U = 66,150 J</em>

Explanation:

<u>Gravitational Potential Energy</u>

Gravitational potential energy is the energy stored in an object because of its vertical position or height in a gravitational field.

It can be calculated with the equation:

U=m.g.h

Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or 9.8\ m/s^2.

The child of mass m=45 Kg is perched above a h=150 m ravine. His gravitational potential energy is:

U=45~Kg\cdot 9.8\ m/s^2\cdot 150\ m

U = 66,150 J

6 0
2 years ago
What do digital signals turn sounds into?
Rashid [163]
Is D) Zeroes and One's :))))
5 0
3 years ago
Read 2 more answers
A 64.0 kg pole vaulter running at 10.2 m/s vaults over the bar. If the vaulter's horizontal component of velocity over the bar i
Lubov Fominskaja [6]

Answer:

h = 5.05 m

Explanation:

given,

mass of pole vaulter, m = 64 Kg

speed of the vaulter,V = 10.2 m/s

horizontal component of velocity in air, v = 1 m/s

height of the jump,h = ?

using energy conservation

     E_i = E_f

\dfrac{1}{2}mv_i^2 + m g h_i= \dfrac{1}{2}mv_f^2+m g h_f

initial height of the vaulter is equal to zero.

\dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2+gh_f

h =\dfrac{v_i^2-v_f^2}{2g}

h =\dfrac{10^2-1^2}{2\times 9.8}

 h = 5.05 m

height of the jump is equal to 5.05 m.

4 0
3 years ago
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