Answer: Receiver, Geometric configuration, accuracy.
Explanation:
Answer: A train traveling initially at 16 m/s is under constant acceleration of 2 m/2. At a distance of 720m it will travel in 20 s, and the final velocity will be 56m/s.
Explanation: To find the answer, we need to know about uniformly accelerated motion.
<h3>
How to solve the problem?</h3>
- We have to find the distance travelled by the train.
- Substituting values, we get,
- We have the equation for final velocity as,
Thus, we can conclude that, a train traveling initially at 16 m/s is under constant acceleration of 2 m/2. At a distance of 720m it will travel in 20 s, and the final velocity will be 56m/s.
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The power P(x) carried by this wave at a point x = P(x) = (μ ω³ A² e⁻²ᵇˣ)/2k
Power time is the pace at which work is completed or energy is delivered; it is expressed as the product of the work completed (W) and the energy transferred (t), or W/t. The variation in the gas pressure ΔP measured from the equilibrium value is also periodic with the same wave number and angular frequency as for the displacement which is given by
ΔP = ΔPmax sin (kx−ωt)
Power is an expression of energy expended through time (effort), of which force is an element, as opposed to force itself, which is the fundamental outcome of an interaction between two objects. Power can be measured and described, but a force is a real physical entity, but power is not. The power is the rate at which the piston is doing work on the element at any instant of time is given by
Power = F ⋅ v
As we mention before in the concept session, the power of the wave is given by
P = ρ ν ω² A s² sin(kx-ωt)
P(x) = 1/2 μ ω² ν A²
= 1/2 μ ω² ω/2 A² e⁻²ᵇˣ
P(x) = (μ ω³ A² e⁻²ᵇˣ)/2k
So, The final answer of power P(x) is P(x) = (μ ω³ A² e⁻²ᵇˣ)/2k.
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Answer:
Its D Kr and Xe
Explanation:
Element from the same group have simmilar properties
If you now the atomic number ،its so easy to know how much electrons they have in outermost shell.
Their speeds will be equal. The one with more mass will have more kinetic energy.