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densk [106]
2 years ago
9

Consider an electron and a proton separated by a distance of 4.5 nm. (a) what is the magnitude of the gravitational force betwee

n them? 5.1128e-20 incorrect: your answer is incorrect. n (b) what is the magnitude of the electric force between them?
Physics
1 answer:
Dmitry_Shevchenko [17]2 years ago
6 0
A.) For letter a, we use the law of universal gravitation using the constant G = 6.674×10−<span>11 m3</span>⋅kg−1⋅s−<span>2

Grav. F = G*m1*m2*(1/d^2)

m1 is mass of electron = </span>9.11 × 10-31<span> kg
m2 is mass of proton = </span>1.67 × 10<span>-27 kg
d = 4.5 nm = 4.5 x 10^-9 m

Grav F = 5.01 x 10^-51 N

b.) </span>For letter b, we use the Coulomb's using the constant k = 9×10^9 N

Electric force = k*Q1*Q2*(1/d^2)
Q1 is charge of electron = -1.6 × 10-19 C
Q2 is charge of proton = +1.6 × 10-19 C

Electric force = 1.14 x 10^-11 N
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The answer is:

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when the tsunami is  a tidal wave which is a series of waves in a water body caused by the displacement of a large volume of water.

7 0
3 years ago
These are equal and opposite forces that do not cause a change in position or motion. True or False
Contact [7]

Answer:

TRUE

Explanation:

Balance forces are usually defined as the two distinct force that acts on an object but in opposite directions. These two acting forces are equal in size or magnitude. When this type of force is applied on any object, it signifies that the object is stationary or it is moving at a constant speed and in the same direction.

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Thus, the above given statement is TRUE.

3 0
3 years ago
two long parrallel wires that are 0.30 m apart carry currents of 5.0 A and 8.0 A in the opposite direction.Find the magnitude of
Vera_Pavlovna [14]

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The force is repulsive.

Explanation: Please see the attachments below

3 0
2 years ago
Air "breaks down" when the electric field strength reaches 3 × 106 n/c, causing a spark. A parallel-plate capacitor is made from
Lera25 [3.4K]

Electric field between the plates of parallel plate capacitor is given as

E = \frac{Q}{A\epsilon_0}

here area of plates of capacitor is given as

A = 0.055 * 0.055

A = 3.025 * 10^{-3}

also the maximum field strength is given as

E = 3 * 10^6 N/C

now we will plug in all data to find the maximum possible charge on capacitor plates

3 * 10^6 = \frac{Q}{3.025 * 10^{-3}*8.85 * 10^{-12}}

Q = 8.03 * 10^{-8} C

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3 0
3 years ago
Read 2 more answers
A fluid in an aquifer is 23.6 m above a reference datum, the fluid pressure (in gage pressure) is 4390 n/m2 and the flow velocit
Phoenix [80]

As per Bernuolli's Theorem total energy per unit mass is given as

\frac{P}{\rho} + \frac{1}{2}v^2 + gH = E

now from above equation

P = 4390 N/m^2

\rho = 0.999 * 10^3

v = 7.22 * 10^{-4} m/s

H = 23.6 m

now by above equation

\frac{4390}{0.999*10^3} + \frac{1}{2}*(7.22*10^{-4})^2 + 9.8*23.6 = E

E = 235.7 J/kg

Part B)

Now energy per unit weight

U = \frac{E}{g}

U = \frac{235.7}{9.8}

U = 24 m

7 0
3 years ago
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