Here we will the speed of seagull which is v = 9 m/s
this is the speed of seagull when there is no effect of wind on it
now in part a)
if effect of wind is in opposite direction then it travels 6 km in 20 min
so the average speed is given by the ratio of total distance and total time
now since effect of wind is in opposite direction then we can say
Part b)
now if bird travels in the same direction of wind then we will have
now we can find the time to go back
Part c)
Total time of round trip when wind is present
now when there is no wind total time is given by
So due to wind time will be more
Answer:
c. 7.5 m/s south
Explanation:
<u>Given the following data;</u>
Velocity of duck = 10m/s due South
Velocity of wind = 2.5m/s
To find the resultant velocity;
Since we know that the duck is flying against a gust of wind, we would have to subtract the velocity of the gust of wind from that of the duck.
This ultimately implies that, the gust of wind (headwind) would decrease the resulting velocity of the duck because it approaches the duck from the front.
Resultant velocity, /V/ = 10 - 2.5
Resultant velocity, /V/ = 7.5m/s South.
<em>Therefore, the resultant velocity of the duck is 7.5m/s south. </em>
Answer:
t = 1.098*RC
Explanation:
In order to calculate the time that the capacitor takes to reach 2/3 of its maximum charge, you use the following formula for the charge of the capacitor:
![Q=Q_{max}[1-e^{-\frac{t}{RC}}]](https://tex.z-dn.net/?f=Q%3DQ_%7Bmax%7D%5B1-e%5E%7B-%5Cfrac%7Bt%7D%7BRC%7D%7D%5D)
(1)
Qmax: maximum charge capacity of the capacitor
t: time
R: resistor of the circuit
C: capacitance of the circuit
When the capacitor has 2/3 of its maximum charge, you have that
Q=(2/3)Qmax
You replace the previous expression for Q in the equation (1), and use properties of logarithms to solve for t:
![Q=\frac{2}{3}Q_{max}=Q_{max}[1-e^{-\frac{t}{RC}}]\\\\\frac{2}{3}=1-e^{-\frac{t}{RC}}\\\\e^{-\frac{t}{RC}}=\frac{1}{3}\\\\-\frac{t}{RC}=ln(\frac{1}{3})\\\\t=-RCln(\frac{1}{3})=1.098RC](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B2%7D%7B3%7DQ_%7Bmax%7D%3DQ_%7Bmax%7D%5B1-e%5E%7B-%5Cfrac%7Bt%7D%7BRC%7D%7D%5D%5C%5C%5C%5C%5Cfrac%7B2%7D%7B3%7D%3D1-e%5E%7B-%5Cfrac%7Bt%7D%7BRC%7D%7D%5C%5C%5C%5Ce%5E%7B-%5Cfrac%7Bt%7D%7BRC%7D%7D%3D%5Cfrac%7B1%7D%7B3%7D%5C%5C%5C%5C-%5Cfrac%7Bt%7D%7BRC%7D%3Dln%28%5Cfrac%7B1%7D%7B3%7D%29%5C%5C%5C%5Ct%3D-RCln%28%5Cfrac%7B1%7D%7B3%7D%29%3D1.098RC)
The charge in the capacitor reaches 2/3 of its maximum charge in a time equal to 1.098RC
Answer:
d= 7.32 mm
Explanation:
Given that
E= 110 GPa
σ = 240 MPa
P= 6640 N
L= 370 mm
ΔL = 0.53
Area A= πr²
We know that elongation due to load given as
A= 42.14 mm²
πr² = 42.14 mm²
r=3.66 mm
diameter ,d= 2r
d= 7.32 mm
