Answer:
The correct answer to the question is (A)
When it hits the heavy rope, compared to the wave on the string, the wave that propagates along the rope has the same (A) frequency
Explanation:
The speed of a wave in a string is dependent on the square root of the tension ad inversely proportional to the square root of the linear density of the string. Generally, the speed of a wave through a spring is dependent on the elastic and inertia properties of the string
Therefore if the linear density of the heavy rope is four times that of light rope the velocity is halved and since
v = f×λ therefore v/2 = f×λ/2
Therefore the wavelength is halved, however the frequency remains the same as continuity requires the frequency of the incident pulse vibration to be transmitted to the denser medium for the wave to continue as the wave is due to vibrating particles from a source for example
Distance is speed x time. Half of the trip is 5.8/2 = 2.9hrs.
640 x 2.9 = 1856mi
580 x 2.9 = 1682mi
1856mi+1682mi=3538mi.
You could also calculate her average speed. This is easy since it was divided in two equal time slices. Average Speed = (640+580)/2 = 610mi/hr
Now 610mi/hr x 5.8hrs = 3538mi
I think its 13...........
Answer:
A) T1 = 269.63 K
T2 = 192.59 K
B) W = -320 KJ
Explanation:
We are given;
Initial volume: V1 = 7 m³
Final Volume; V2 = 5 m³
Constant Pressure; P = 160 KPa
Mass; m = 2 kg
To find the initial and final temperatures, we will use the ideal gas formula;
T = PV/mR
Where R is gas constant of helium = R = 2.0769 kPa.m/kg
Thus;
Initial temperature; T1 = (160 × 7)/(2 × 2.0769) = 269.63 K
Final temperature; T2 = (160 × 5)/(2 × 2.0769) = 192.59 K
B) world one is given by the formula;
W = P(V2 - V1)
W = 160(5 - 7)
W = -320 KJ
The power dissipated across a component can be calculated through the formula P=I^2xR
Substituting the values in we get P=(0.5)^2x10=2.5W
