Answer:
1.23917 m
0.14323 s
Explanation:
v = Speed of sound in dry air at 20 °C = 343.5 m/s
f = Frequency of note C# = 277.2 /s = 277.2 Hz
λ = Wavelength

Wavelength = 1.23917 m
Distance the wave needs to travel is 49.2 m
Time = Distance / Speed

Time taken for the sound to travel across the concert hall is 0.14323 s
<span>Based
on Newton's law of universal gravitation, the equation for the
gravitational force exerted by an object on another object is given by:
F = Gm1m2/(r^2)
where G is the universal gravitational constant, F is the gravitational
force exerted, m1 is the mass of the first object, m2 is the mass of the
second object, and r is the separation distance between the two
objects.
If the mass of both objects were doubled, then we would have: m1' * m2' =
(2m1) * (2m2) = 4m1m2. Assuming r stays constant (G is a constant so
that won't change anyway), then this means that the new force will be 4
times greater, ie 8N * 4 = 32N of gravitational force. </span>
Answer:
right A, B, C, D
Explanation:
They ask which statements are true
A) Right. The decrease in amplitude is due to the dissipation of energy by friction and is called damping
B) Right. In resonant processes the amplitude of the oscillation increases, being a forced oscillation
C) Right. In a system with energy loss, the amplitude must decrease, therefore energy must be supplied to compensate for the loss.
D) Right. It is a resonant process the driving force keeps the oscillation of the system
Answer:633 m
Explanation:
First we have moved 300 m in North
let say it as point a and its vector is 
after that we have moved 500 m northeast
let say it as point b
therefore position of b with respect to a is
r
Therefore position of b w.r.t to origin is


![r_b=500cos(45)\hat{i}+\left [ 250\sqrt{2}+300\right ]\hat{j}](https://tex.z-dn.net/?f=r_b%3D500cos%2845%29%5Chat%7Bi%7D%2B%5Cleft%20%5B%20250%5Csqrt%7B2%7D%2B300%5Cright%20%5D%5Chat%7Bj%7D)
after this we moved 400 m
south of east i.e.
below from positive x axis
let say it as c


![r_c=500cos(45)\hat{i}+\left [ 250\sqrt{2}+300\right ]\hat{j}+400cos(60)\hat{i}-400sin(60)\hat{j}](https://tex.z-dn.net/?f=r_c%3D500cos%2845%29%5Chat%7Bi%7D%2B%5Cleft%20%5B%20250%5Csqrt%7B2%7D%2B300%5Cright%20%5D%5Chat%7Bj%7D%2B400cos%2860%29%5Chat%7Bi%7D-400sin%2860%29%5Chat%7Bj%7D)
![r_c=\left [ 250\sqrt{2}+200\right ]\hat{i}+\left [ 250\sqrt{2}+300-200\sqrt{3}\right ]\hat{j}](https://tex.z-dn.net/?f=r_c%3D%5Cleft%20%5B%20250%5Csqrt%7B2%7D%2B200%5Cright%20%5D%5Chat%7Bi%7D%2B%5Cleft%20%5B%20250%5Csqrt%7B2%7D%2B300-200%5Csqrt%7B3%7D%5Cright%20%5D%5Chat%7Bj%7D)
magnitude is ![\sqrt{\left [ 250\sqrt{2}+200\right ]^2+\left [ 250\sqrt{2}+300-200\sqrt{3}\right ]^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cleft%20%5B%20250%5Csqrt%7B2%7D%2B200%5Cright%20%5D%5E2%2B%5Cleft%20%5B%20250%5Csqrt%7B2%7D%2B300-200%5Csqrt%7B3%7D%5Cright%20%5D%5E2%7D)
=633.052
for direction

with x -axis
Answer:

Explanation:
If the stone will reach the top position of flag pole at t = 0.5 s and t = 4.1 s
so here the total time of the motion above the top point of pole is given as

now we have



so this is the speed at the top of flag pole
now we have



now the height of flag pole is given as


