Science is an art, u cant make art without creativity
Answer:
The true course:
north of east
The ground speed of the plane: 96.68 m/s
Explanation:
Given:
= velocity of wind = ![30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h](https://tex.z-dn.net/?f=30%5C%20km%2Fh%5C%20S45%5E%5Ccirc%20E%20%3D%20%2830%5Ccos%2045%5E%5Ccirc%5C%20%5Chat%7Bi%7D-30%5Csin%2045%5E%5Ccirc%5C%20%5Chat%7Bj%7D%29%5C%20km%2Fh%20%3D%20%2821.21%5C%20%5Chat%7Bi%7D-21.21%5C%20%5Chat%7Bj%7D%29%5C%20km%2Fh)
= velocity of plane in still air = ![100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h](https://tex.z-dn.net/?f=100%5C%20km%2Fh%5C%20N60%5E%5Ccirc%20E%20%3D%20%28100%5Ccos%2060%5E%5Ccirc%5C%20%5Chat%7Bi%7D%2B100%5Csin%2060%5E%5Ccirc%5C%20%5Chat%7Bj%7D%29%5C%20km%2Fh%20%3D%20%2850%5C%20%5Chat%7Bi%7D%2B86.60%5C%20%5Chat%7Bj%7D%29%5C%20km%2Fh)
Assume:
= resultant velocity of the plane
= direction of the plane with the east
Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.
![\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h](https://tex.z-dn.net/?f=%5Ctherefore%20V_r%20%3D%20V_p%2BV_w%5C%5C%5CRightarrow%20V_r%20%3D%20%2850%5C%20%5Chat%7Bi%7D%2B86.60%5C%20%5Chat%7Bj%7D%29%5C%20km%2Fh%2B%2821.21%5C%20%5Chat%7Bi%7D-21.21%5C%20%5Chat%7Bj%7D%29%5C%20km%2Fh%5C%5C%5CRightarrow%20V_r%20%3D%20%2871.21%5C%20%5Chat%7Bi%7D%2B65.39%5C%20%5Chat%7Bj%7D%29%5C%20km%2Fh)
Let us find the direction of this resultant velocity with respect to east direction:
![\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Ctan%5E%7B-1%7D%28%5Cdfrac%7B65.39%7D%7B71.21%7D%29%5C%5C%5CRightarrow%20%5Ctheta%20%3D%2040.29%5E%5Ccirc)
This means the the true course of the plane is in the direction of
north of east.
The ground speed will be the magnitude of the resultant velocity of the plane.
![\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h](https://tex.z-dn.net/?f=%5Ctherefore%20Magnitude%20%3D%20%5Csqrt%7B71.21%5E2%2B65.39%5E2%7D%20%3D%2096.68%5C%20km%2Fh)
Hence, the ground speed of the plane is 96.68 km/h.
Answer: Drink water, practice, do some light stretches
Explanation:
The spider will cross the driveway <u>25.71 in seconds</u>.
Why?
It's a conversion problem, so, in order to solve it, we need to convert the units of the given information.
Let's convert the given speed (in cm/s) to m/s.
We know that:
![1m=100cm](https://tex.z-dn.net/?f=1m%3D100cm)
So, converting we have:
![14\frac{cm}{s}*\frac{1m}{100cm}=0.14\frac{m}{s}](https://tex.z-dn.net/?f=14%5Cfrac%7Bcm%7D%7Bs%7D%2A%5Cfrac%7B1m%7D%7B100cm%7D%3D0.14%5Cfrac%7Bm%7D%7Bs%7D)
Then, calculating the time, we have:
![distance=speed*time\\\\3.6m=0.14\frac{m}{s}*time\\\\time=\frac{3.6m}{0.14\frac{m}{s}}=25.71s](https://tex.z-dn.net/?f=distance%3Dspeed%2Atime%5C%5C%5C%5C3.6m%3D0.14%5Cfrac%7Bm%7D%7Bs%7D%2Atime%5C%5C%5C%5Ctime%3D%5Cfrac%7B3.6m%7D%7B0.14%5Cfrac%7Bm%7D%7Bs%7D%7D%3D25.71s)
We have that the spider will cross the driveway in 25.71 seconds.
Have a nice day!