Answer:
A) The wave equation is given as
According to the above equation, k = 0.004 and ω = 5.05.
Using the following identities, we can find the period of the wave.
T = 1.25 s.
For the horizontal distance travelled by one period of time, x = λ.
B) The wave number, k = 0.004 . The number of waves per second is the frequency, so f = 0.8.
C) The propagation speed of the wave is
The velocity of the wave is the derivative of the position function.
The maximum velocity is when the derivative of the velocity function is equal to zero.
In order this to be zero, cosine term must be equal to zero.
The reason that cosine term is set to be 5π/2 is that time cannot be zero. For π/2 and 3π/2, t<0.
Answer:
Driving in a straight line at 60 miles per hour
Explanation:
In the first case there's an acceleration that modifies the direction of the movement.
In the second case there's a lineal acceleration that increases the speed of the car.
in the third case there's a negative acceleration that reduces the speed of the car.
On the third case the speed is constant so the acceleration is 0 mi/s^2
Answer:
magnetic field will allow the electron to go through 2 x 
T k
Explanation:
Given data in question
velocity = 5.0 × 
electric filed =
To find out
what magnetic field will allow the electron to go through, undeflected
solution
we know if electron move without deflection i.e. net force is zero on electron and we can say both electric and magnetic force equal in magnitude and opposite in directions
so we can also say
F(net) = Fe + Fb i.e. = 0
q V B + q E = 0
q will be cancel out
+ 5e + 7i × B = 0
B = 2 x T k
The speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.
<h3>
Speed of the satellite</h3>
v = √GM/r
where;
- M is mass of Earth
- G is universal gravitation constant
- r is distance from center of Earth = Radius of earth + 4930 km
v = √[(6.626 x 10⁻¹¹ x 5.97 x 10²⁴) / ((6371 + 4930) x 10³)]
v = 5,916.36 m/s
Thus, the speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.
Learn more about speed here: brainly.com/question/6504879
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Answer:
current in loops is 52.73 μA
Explanation:
given data
side of square a = b = 2.40 cm = 0.024 m
resistance R = 1.20×10^−2 Ω
edge of the loop c = 1.20 cm = 0.012 m
rate of current = 120 A/s
to find out
current in the loop
solution
we know current formula that is
current = voltage / resistance .................a
so current = 1/R × d∅/dt
and we know here that
flux ∅ = ( μ×I×b / 2π ) × ln (a+c/c) ...............b
so
d∅/dt = ( μ×b / 2π ) × ln (a+c/c) × dI/dt ...........c
so from equation a we get here current
current = ( μ×b / 2πR ) × ln (a+c/c) × dI/dt
current = ( 4π×
×0.024 / 2π(1.20×
) × ln (0.024 + 0.012/0.012) × 120
solve it and we get current that is
current = 4 ×× 1.09861 × 120
current = 52.73 ×
A
so here current in loops is 52.73 μA
