Ans: Beat Frequency = 1.97HzExplanation:
The fundamental frequency on a vibrating string is

<span> -- (A)</span>
<span>here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg </span>
Plug in the values in Equation (A)
<span>so </span>

<span> = 197.97Hz </span>
<span>the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
-i</span>
Answer: Part 1: Propellant Fraction (MR) = 8.76
Part 2: Propellant Fraction (MR) = 1.63
Explanation: The Ideal Rocket Equation is given by:
Δv = 
Where:
is relationship between exhaust velocity and specific impulse
is the porpellant fraction, also written as MR.
The relationship
is: 
To determine the fraction:
Δv = 

Knowing that change in velocity is Δv = 9.6km/s and
= 9.81m/s²
<u>Note:</u> Velocity and gravity have different measures, so to cancel them out, transform km in m by multiplying velocity by 10³.
<u />
<u>Part 1</u>: Isp = 450s

ln(MR) = 
ln (MR) = 2.17
MR = 
MR = 8.76
<u>Part 2:</u> Isp = 2000s

ln (MR) = 
ln (MR) = 0.49
MR = 
MR = 1.63
Time=speed/acceleration
Gravitaional Acceleration=9.8 m/s^2
Speed=24.5 m/s
Time=24.5/9.8=2.5 s
The value of the swimmer's power output is calculated by dividing the work done by the time it took for the work to be completed. From the given in this item,
P = 3560 J/ 55 s = 64.73 W
Rounding off to two significant figures will give us 65 W.
Answer:
<em>0.45 mm</em>
Explanation:
The complete question is
a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?
A) 0.45 mm
B) 0.63 mm
C.) 0.68 mm
D) 0.91 mm
Current in the fuse is 1.0 A
Current density of the fuse when it melts is 620 A/cm^2
Area of the wire in the fuse = I/ρ
Where I is the current through the fuse
ρ is the current density of the fuse
Area = 1/620 = 1.613 x 10^-3 cm^2
We know that 10000 cm^2 = 1 m^2, therefore,
1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2
Recall that this area of this wire is gotten as
A = 
where d is the diameter of the wire
1.613 x 10^-7 = 
6.448 x 10^-7 = 3.142 x 
=
d = 4.5 x 10^-4 m = <em>0.45 mm</em>