Answer:
They will run parallel to each other as the none of a straight pole cannot be bent in such a way where one side can turn without the other turning.
Answer:
53.895 m.
Explanation:
Using the equation of motion,
v² = u² + 2as .............. Equation 1
Where v = final velocity of the swan, u = initial velocity of the swan, a = acceleration of the swan, s = distance covered by the swan.
make s the subject of the equation,
s = (v² - u²)/2a----------- Equation 2
Given: v = 6.4 m/s, u = 0 m/s ( from rest) a = 0.380 m/s².
Substitute into equation 2
s = (6.4²-0²)/(2×0.380)
s = 40.96/0.76
s = 53.895 m.
Hence the swan will travel 53.895 m before becoming airborne.
Answer:
Question 1 the anwser is 300 meters. Question 2 the answer is speed.
is Explanation:
Answer:
We have learned that refraction occurs as light passes across the boundary between two media. Refraction is merely one of several possible boundary behaviors by which a light wave could behave when it encounters a new medium or an obstacle in its path.
Answer:
18.7842493212 W
Explanation:
T = Tension = 1871 N
= Linear density = 3.9 g/m
y = Amplitude = 3.1 mm
= Angular frequency = 1203 rad/s
Average rate of energy transfer is given by
![P=\dfrac{1}{2}\sqrt{T\mu}\omega^2y^2\\\Rightarrow P=\dfrac{1}{2}\sqrt{1871\times 3.9\times 10^{-3}}\times 1203^2\times (3.1\times 10^{-3})^2\\\Rightarrow P=18.7842493212\ W](https://tex.z-dn.net/?f=P%3D%5Cdfrac%7B1%7D%7B2%7D%5Csqrt%7BT%5Cmu%7D%5Comega%5E2y%5E2%5C%5C%5CRightarrow%20P%3D%5Cdfrac%7B1%7D%7B2%7D%5Csqrt%7B1871%5Ctimes%203.9%5Ctimes%2010%5E%7B-3%7D%7D%5Ctimes%201203%5E2%5Ctimes%20%283.1%5Ctimes%2010%5E%7B-3%7D%29%5E2%5C%5C%5CRightarrow%20P%3D18.7842493212%5C%20W)
The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W