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AfilCa [17]
3 years ago
14

A proton is located at the point (x = 1.0 nm, y = 0.0 nm) and an electron is located at the point (x = 0.0 nm, y = 4.0 nm). Find

the magnitude of the electrostatic force that each one exerts on the other. (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2, e = 1.6 × 10-19 C) A) 5.3 × 10-18 NB) 5.3 × 108 N C) 1.4 × 10-11 N D) 5.9 × 10-15 N
Physics
1 answer:
enot [183]3 years ago
7 0

Answer:

C : 1.4*10^(-11) N.

Explanation:

q_1 = q_2 = 1.6 * 10^(-19)

R^2 = (1)^2 + (4)^2 = 1.7 * 10^(-17) m^2

The coulomb's law is as follows:

F_e = k*q_1*q_2 / R^2

F_e = k*q^2 / R^2

F_e = (9.0*10^9) * (1.6 * 10^(-19))^2 / 1.7 * 10^(-17)

F_e = 1.35 * 10^(-11) N

Hence, answer closes to obtained is option C : 1.4*10^(-11) N.

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the train accelerates from 30 km/h to 45 km/h in 15 secs. a. find its acceleration. b. distance it travels during this time ...
Sonbull [250]
Acceleration = v-u/t
= (45-30)/15
= 1 km/h


Distance = ut + 1/2 at^2
s = [30 x 15]  + 1/2 x 1 x 15^2 
= 562.5 km

Hope this helps

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3 years ago
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What is the difference between acceleration due to gravity and gravitational force?​
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Answer:

Gravity.” or “Gravity causes the moon to orbit the earth.” “Acceleration due to gravity” is much more precise. It is the acceleration that an object experiences when it is dropped with no other external forces on it. ... Near the surface of the earth the acceleration due to gravity is about 9.8

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What effect does plate movements have on geological events​
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8 0
4 years ago
Light of wavelength 648.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 57.5 cm from the slit.
boyakko [2]

Answer:

71.0 \mu m

Explanation:

The formula for the single-slit diffraction is

y=\frac{n\lambda D}{d}

where

y is the distance of the n-minimum from the centre of the diffraction pattern

D is the distance of the screen from the slit

d is the width of the slit

\lambda is the wavelength of the light

In this problem,

\lambda=648.0 nm=6.48\cdot 10^{-7}m

D=57.5 cm=0.575 m

y=1.05 cm=0.0105 m, with n=2 (this is the distance of the 2nd-order minimum from the central maximum)

Solving the formula for d, we find:

d=\frac{n\lambda D}{y}=\frac{2(6.48\cdot 10^{-7} m)(0.575 m)}{0.0105 m}=7.10\cdot 10^{-5} m= 71.0 \mu m

6 0
3 years ago
A group of hikers hear an echo 2.6 s after they shout. The temperature is 20 ◦C. How far away is the mountain that reflected the
Vadim26 [7]

Answer:

x = 445.9\,m

Explanation:

The velocity of sound in air at 20 °C is 343 meters per second. The distance of mountain can be derived by the following expression:

2\cdot x = v_{air}\cdot \Delta t

x = \frac{v_{air}\cdot \Delta t}{2}

x = \frac{(343\,\frac{m}{s} )\cdot (2.6\,s)}{2}

x = 445.9\,m

4 0
4 years ago
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