The "ringing sound" are actually the soundwaves travelling from the phone to your ears.
Answer:
The speed of the wave on the longer wire is 580m/s
Explanation:
The velocity possessed by a stretched string is directly proportional to the tension in the string and inversely proportional to the mass per unit length of the string. Mathematically,
V = √T/m
Where V is the velocity of wave in the wire
T is the tension in the wire
M is the mass per unit length of the wire
Let m1 and m2 be the mass per unit length of the wires
Let T1 and T2 be their respective tensions
Since the tension and mass of the wire is the same
m1= m2= m.
T1=T2=T
Let m1 =M/l
m2 =M/4l( since the second is tour times as far apart)
V1 = 290m/s(velocity in shorter wire)
V2 is the velocity of the longer wire.
V1 = √T/(m/l)
290 = √Tl/m
290² = Tl/m... 1
V2 = √4Tl/m
V2²= 4Tl/m... 2
Dividing equation 1 by 2 we have;
290²/V2² = {Tl/m}/{4Tl/m}
290²/V2² = Tl/m × m/4Tl
290²/V2² = 1/4
Cross multiplying we have;
V2² = 290²×4
V2 = √290²×4
V2 = 290×2
V2 = 580m/s
Answer: C = Q/4πR
Explanation:
Volume(V) of a sphere = 4πr^3
Charge within a small volume 'dV' is given by:
dq = ρ(r)dV
ρ(r) = C/r^2
Volume(V) of a sphere = 4/3(πr^3)
dV/dr = (4/3)×3πr^2
dV = 4πr^2dr
Therefore,
dq = ρ(r)dV ; dq =ρ(r)4πr^2dr
dq = C/r^2[4πr^2dr]
dq = 4Cπdr
FOR TOTAL CHANGE 'Q', we integrate dq
∫dq = ∫4Cπdr at r = R and r = 0
∫4Cπdr = 4Cπr
Q = 4Cπ(R - 0)
Q = 4CπR - 0
Q = 4CπR
C = Q/4πR
The value of C in terms of Q and R is [Q/4πR]
The projectile has a height <em>h</em> at time <em>t</em> given by
<em>h</em> = (14.0 m/s) <em>t</em> - 1/2 <em>g t</em> ²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for <em>t</em> when <em>h</em> = 0 :
0 = (14.0 m/s) <em>t</em> - 1/2 <em>g t</em> ²
0 = 1/2 <em>t</em> (28.0 m/s - <em>g t</em> )
1/2 <em>t</em> = 0 <u>or</u> 28.0 m/s - <em>g</em> <em>t</em> = 0
The first equation says <em>t</em> = 0, which refers to the moment the gun is first fired, so we ignore that solution. We're left with
28.0 m/s - <em>g t</em> = 0
<em>t</em> = (28.0 m/s) / <em>g</em>
<em>t</em> = (28.0 m/s) / (9.80 m/s²)
<em>t</em> ≈ 2.86 s
Answer:
I think it's B: b.
Use the number that lines up with the bottom part of the blue line on the left of the thermometer
Explanation:
The liquid in the thermometer makes a "U" crescent and you have to use the bottom of the "U" touches the the lines on the side.