117 m/sec is the speed of a transverse wave in a rope of length 3. 1 m and mass 86 g under a tension of 380 n.
The wave speed v is given by
v= √τ/μ
where τ is the tension in the rope and μ is the linear mass density of the rope.
The linear mass density is the mass per unit length of rope :
μ= m / L = (0.086 kg)/(3.1 m)=0.0277 kg/m.
v= 
= 117.125 m/sec (approx. 117 m/sec
In physics, a transverse wave is a wave whose oscillations are perpendicular to the direction of the wave's advance. This is in contrast to a longitudinal wave which travels in the direction of its oscillations. Water waves are an example of transverse wave.
Transverse waves commonly occur in elastic solids due to the shear stress generated; the oscillations in this case are the displacement of the solid particles away from their relaxed position, in directions perpendicular to the propagation of the wave. These displacements correspond to a local shear deformation of the material. Hence a transverse wave of this nature is called a shear wave. Since fluids cannot resist shear forces while at rest, propagation of transverse waves inside the bulk of fluids is not possible.
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Answer:
The correct answer is a
Explanation:
At projectile launch speeds are
X axis vₓ = v₀ = cte
Y axis 
= v_{oy} –gt
The moment is defined as
p = mv
For the x axis
pₓ = mvₓ = m v₀ₓ
As the speed is constant the moment is constant
For the y axis
p_{y} = m v_{y} = m (v_{oy} –gt) = m v_{oy} - m (gt)
Speed changes over time, so the moment also changes over time
Let's examine the answer
i True
ii False. The moment changes with time
The correct answer is a
The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance 
= 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 
V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q = 
Therefore, substitute the values into the above formula as follows.
Q =
= 
= 
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is .
Answer:
a) Θ = ω₀*t + ½αt² To complete first revolution 2π rads = 0*t + ½αt² and to complete the first and second combined 4π rads = 0*t + ½α(t+0.810s)² Divide second by first: 2 = (t + 0.810s)² / t² This is quadratic in t and has roots at t = -0.336 s ← ignore and t = 1.96 s ◄ b) Use either equation from above: 2π rads = 0*t + ½α(1.96s)² α = 3.27 rad/s² ◄ Hope this helps!
Explanation:
