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Scorpion4ik [409]
3 years ago
11

2 Points

Physics
1 answer:
juin [17]3 years ago
3 0
Answer: c




Explanation:

Analytical methods

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In the Bohr model of the hydrogen atom, an electron({rm mass};m=9.1; times 10^{ - 31;}{rm kg}) orbits a proton at a distance of
max2010maxim [7]

Answer:

n=6.56×10¹⁵Hz

Explanation:

Given Data

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

To find

Revolutions per second

Solution

Let F be the force of attraction

let n  be the number of revolutions per sec made by the electron around the nucleus then the centripetal force is given by

F=mω²r......................where ω=2π  n

F=m4π²n²r...............eq(i)

as the values given where

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

we have to find n from eq(i)

n²=F/(m4π²r)

n^{2} =\frac{8.2*10^{-8} }{9.11*10^{-31}* 4\pi^{2} *5.3*10^{-11}  }\\ n^{2}=4.31*10^{31}\\ n=\sqrt{4.31*10^{31}}\\ n=6.56*10^{15}Hz

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3 years ago
Calculate the kinetic energy of Earth due to its spinning about its axis. Krot = J Compare your answer with the kinetic energy o
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Answer:

Explanation:

Answer: Let ke = 1/2 IW^2 = 1/2 kMr^2 W^2 be Earth's rotational KE. W = 2pi/24 radians per hour rotation speed and k = 2/5 for a solid sphere M is Earth mass, r = 6.4E6 m.

Then ke = 1/2 2/5 6E24 (6.4E6)^2 (2pi/(24*3600))^2 = ? Joules. You can do the math, note W is converted to radians per second for unit consistency.

Let KE = 1/2 KMR^2 w^2 be Earth's orbital KE. w = 2pi/(365*24) radians per hour K = 1 for a point mass. Note I used 365 days, a more precise number is 365.25 days per year, which is why we have Leap Years.

Find KE/ke = 1/2 KMR^2 w^2//1/2 kMr^2 W^2 = (K/k)(w/W)^2 (R/r)^2 = (5/2) (365)^2 (1.5E11/6.4E6)^2 = 7.81E9 ANS

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As a cat pounces on a mouse, its muscles consume 10 units of potential energy (which the cat previously gained from the food it
taurus [48]
6 units of energy were transferred into heat.
7 0
3 years ago
The acceleration of the body in the graph given below is____
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Acceleration = -12 / 8 = - .... m/s^2
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19. What is shape of BF3​
iren2701 [21]

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The geometry of the BF 3 molecule is called trigonal planar (see Figure 5). The fluorine atoms are positioned at the vertices of an equilateral triangle. The F-B-F angle is 120° and all four atoms lie in the same plane.

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