Answer:
Explanation:
Given
mass (m)= 20 kg
acceleration (a)= 10 m/s^2
Force (f)= m a
= 20 * 10
= 200 N
The equation has one zero, and the system has three solutions ( 1 real and 2 complex solutions).
<h3>
Solution of the polynomial equation</h3>
The solution of the polynomial equation is determined as follows;
3x³ + x = 2x² + 1
3x³ - 2x² + x - 1 = 0
Thus, we can conclude that the equation has one zero (real solution), and the system has three solutions ( 1 real and 2 complex solutions).
Learn more about polynomial equations here: brainly.com/question/2833285
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Answer:
a) 
is the speed of each proton
b) 
Explanation:
Given:
radius of path of motion, 
we know charge on protons, 
magnetic field strength, 
we've mass of proton, 
a)
From the equivalence of magnetic force and the centripetal force on the proton:
where:
v = speed of the proton
is the speed of each proton
b)
Now the centripetal force on each proton:
<span>Here the force that is applied between the electron and proton is centripetal, so equate the two forces to determine the velocity.
We know charge of the electron which for both Q1 and Q2, e = 1.60 x 10^-19 C
The Coulombs Constant k = 9.0 x 10^9
Radius r = 0.053 x 10^-9m = 5.3 x 10^-11 m
Mass of the Electron = 9.11 x 10^-31
F = k x Q1 x Q2 / r^2 = m x v^2 / r(centripetal force)
ke^2 / r^2 = m x v^2 / r => v^2 = ke^2 / m x r
v^2 = ((1.60 x 10^-19)^2 x 9.0 x 10^9) / (9.11 x 10^-31 x 5.3 x 10^-11 )
v^2 = 4.77 x 10^12 = 2.18 x 10^6 m/s
Since one orbit is the distance,
one orbit = circumference = 2 x pi x r; distance s = v x t.
v x t = 2 x pi x r => t = (2 x 3.14 x 5.3 x 10^-11) / (2.18 x 10^6)
t = 33.3 x 10^-11 / 2.18 x 10^6 = 15.27 x 10^-17 s
Revolutions per sec = 1 / t = 1 / 15.27 x 10^-17 = 6.54 x 10^15 Hz</span>
Answer:
1. True WA > WB > WC
Explanation:
In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.
A) Work is the product of force by distance and the cosine of the angle between them
WA = W h cos 0
WA = mg h
B) On a ramp without rubbing
Sin30 = h / L
L = h / sin 30
WB = F d cos θ
WB = F L cos 30
WB = mf (h / sin30) cos 30
WB = mg h ctan 30
C) Ramp with rubbing
W sin 30 - fr = ma
N- Wcos30 = 0
W sin 30 - μ W cos 30 = ma
F = W (sin30 - μ cos30)
WC = mg (sin30 - μ cos30) h / sin30
Wc = mg (1 - μ ctan30) h
When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum
Let's review the claims
1. True The work of gravity is the greatest and the work where there is friction is the least
2 False. The job where there is friction is the least
3 False work with rubbing is the least
4 False work with rubbing is the least
