Why does a torque rod have a stronger magnetic field than a torque coil?
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The average speed will be 2.38×10⁶ m/sec.The average speed of an object indicates the pace at which it will traverse a distance. The metric unit of speed is the meter per second.
<h3>What is the average speed?</h3>The total distance traveled by an object divided by the total time taken is the average speed.
The speed calculated at any particular instant of time is known as the instantaneous speed.
Given data;
Distance travelled = 4.12x10¹⁶ meter
Time period= 1.73x10¹⁰ sec
The average speed is found as
Hence, the average speed will be 2.38×10⁶ m/sec.
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Answer:
(a)
(b)
Explanation:
Parameter given:
Electric field, E =
(a) Electric force is given (in terms of electric field) as a product of electric charge and electric field.
Mathematically:
Electric charge, q, of an electron =
(b) This electrostatic force causes the electron to accelerate with an equivalent force:
F = -ma
where m = mass of an electron
a = acceleration of electron
(Note: the force is negative cos the direction of the force is opposite the direction of the electron)
Therefore:
Mass, m, of an electron =
=>
The acceleration of the electron is
Answer:
The answer to the question
The steady state response is u₂(t) = -cos(3t + π/4)
of the form R·cos(ωt−δ) with R = , ω = 3 and δ = -π/4
Explanation:
To solve the question we note that the equation of motion is given by
m·u'' + γ·u' + k·u = F(t) where
m = mass = 2.00 kg
γ = Damping coefficient = 1
k = Spring constant = 3 N·m
F(t) = externally applied force = 27·cos(3·t)−18·sin(3·t)
Therefore we have 2·u'' + u' + 3·u = 27·cos(3·t)−18·sin(3·t)
The homogeneous equation 2·u'' + u' + 3·u is first solved as follows
2·u'' + u' + 3·u = 0 where putting the characteristic equation as
2·X² + X + 3 = 0 we have the solution given by =
This gives the general solution of the homogeneous equation as
u₁(t) =
For a particular equation of the form 2·u''+u'+3·u = 27·cos(3·t)−18·sin(3·t) which is in the form u₂(t) = A·cos(3·t) + B·sin(3·t)
Then u₂'(t) = -3·A·sin(3·t) + 3·B·cos(3·t) also u₂''(t) = -9·A·cos(3·t) - 9·B·sin(3·t) from which 2·u₂''(t)+u₂'(t)+3·u₂(t) = (3·B-15·A)·cos(3·t) + (-3·A-15·B)·sin(3·t). Comparing with the equation 27·cos(3·t)−18·sin(3·t) we have
3·B-15·A = 27
3·A +15·B = 18
Solving the above linear system of equations we have
A = -1.5, B = 1.5 and u₂(t) = A·cos(3·t) + B·sin(3·t) becomes 1.5·sin(3·t) - 1.5·cos(3·t)
u₂(t) = 1.5·(sin(3·t) - cos(3·t) = ·cos(3·t + π/4)
The general solution is then u(t) = u₁(t) + u₂(t)
however since u₁(t) = ⇒ 0 as t → ∞ the steady state response = u₂(t) =
·cos(3·t + π/4) which is of the form R·cos(ωt−δ) where
R =
ω = 3 and
δ = -π/4