Answer:
245.45km in a direction 21.45° west of north from city A
Explanation:
Let's place the origin of a coordinate system at city A.
The final position of the airplane is given by:
rf = ra + rb + rc where ra, rb and rc are the vectors of the relative displacements the airplane has made. If we separate this equation into its x and y coordinates:
rfX = raX+ rbX + rcX = 175*cos(30)-150*sin(20)-190 = -89.75km
rfY = raY + rbY + rcT = 175*sin(30)+150*cos(20) = 228.45km
The module of this position is:
And the angle measure from the y-axis is:
So the answer is 245.45km in a direction 21.45° west of north from city A
Answer:
Explanation:
Initial separation of plate = d
final separation = 2d
The capacitance of the capacitor will reduce from C to C/2 because
capacitance = ε A / d
d is distance between plates.
As the batteries are disconnected , charge on the capacitor becomes fixed .
Initial charge on the capacitor
= Capacitance x potential difference
Q = C ΔV
Final charge will remain unchanged
Final charge = C ΔV
Final capacitance = C/2
Final potential difference = charge / capacitance
= C ΔV / C/2
= 2 ΔV
Potential difference is doubled after the pates are further separated.
><span>It can travel through vacuum.
The rays must travel in the vacuum of space between Earth's atmosphere and the sun.</span>
0N. The net force acting on this firework is 0.
The key to solve this problem is using the net force formula based on the diagram shown in the image. Fnet = F1 + F2.....Fn.
Based on the free-body diagram, we have:
The force of gases is Fgases = 9,452N
The force of the rocket Frocket = -9452
Then, the net force acting is:
Fnet = Fgases + Frocket
Fnet = 9,452N - 9,452N = 0N
