Good coding specialists share which of the following characteristics?
2 answers:
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Answer:
E = 440816.32 N/C
Explanation:
Given data:
Three point charge of charge equal to +3.0 micro coulomb
fourth point charge = - 3.0 micro coulomb
side of square = 0.50 m
N.m^2/c^2
Due to having equal charge on center of square, 2 charge produce equal electric field at center and other two also produce electric field at center of same value
So we have
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plugging all value
E = 440816.32 N/C
Refer to the diagram shown.
When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N
By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m
When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.
If damping is ignored, the equation of motion is
F = m * acceleration
or
Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)
x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0
The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32
The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.
Answer:
0.32 m (single amplitude), or
0.64 m (double amplitude)
When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N
By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m
When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.
If damping is ignored, the equation of motion is
F = m * acceleration
or
Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)
x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0
The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32
The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.
Answer:
0.32 m (single amplitude), or
0.64 m (double amplitude)
Answer:
a) is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.
b) is speed of the rocket going when it stops accelerating.
c)
d)
e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.
Explanation:
Given:
acceleration of rocket,
time for which the rocket accelerates,
<u>For the course of upward acceleration:</u>
using eq. of motion,
where:
initial velocity of the rocket at the launch
height the rocket travels just before its fuel finishes off
so,
a) is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.
<u>Now the velocity of the rocket just after the fuel is finished:</u>
b) is speed of the rocket going when it stops accelerating.
After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.
Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:
using equation of motion,
where:
acceleration due to gravity
final velocity of the rocket at the top height
c) So the total height at which the rocket gets:
d)
Time taken by the rocket to reach the top height after the fuel is over:
Now the time taken to fall from the total height:
Hence the total time taken by the rocket to strike back on the earth:
e)
Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.