4. gravitational
5.all
6.sound
7.all
8.chemical
9.electrical.
10.sound
Answer:
3743.489 kg
Explanation:
F_g = 591 N
G = 6.674x10^-11 constant of gravity
m_1 = 95 kg
m_2 = unknown
r = 4990*1000 =
F_g = G[(m_1*m_2)/r^2]
591 N = 6.674x10^-11[(95*m_2)/4990^2]
8.855 = [(95*m_2)/4990^2]
355631.472 = 95*m_2
m_2 = 3743.489 kg
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