Explanation:
Solution:
Let the time be
t1=35min = 0.58min
t2=10min=0.166min
t3=45min= 0.75min
t4=35min= 0.58min
let the velocities be
v1=100km/h
v2=55km/h
v3=35km/h
a. Determine the average speed for the trip. km/h
first we have to solve for the distance
S=s1+s2+s3
S= v1t1+v2t2+v3t3
S= 100*0.58+55*0.166+35*0.75
S=58+9.13+26.25
S=93.38km
V=S/t1+t2+t3+t4
V=93.38/0.58+0.166+0.75+0.58
V=93.38/2.076
V=44.98km/h
b. the distance is 93.38km
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
Correct answer: B
The force of gravity acting on the object on the moon can be found by multiplying its mass by the acceleration due to gravity on the moon. The acceleration due to gravity on the moon is a constant and is approximately 1.63m/s².
120kg×1.63m/s²=195.6kg.m/s²
kg.m/s²=N
<span>The force of gravity acting on the object on the moon would be of approximately 196N.</span>
6.
A. 1575 - 1265 = 310J
B. KE 1/2 MV^2
V=√2·KE/M = √2(310)/12 V = 7.2mls
C. PE = 1265 = mgh
h= 1265/mg = 1265/(12)(92) h= 10.8m
7.
A. KE = 1/2 mv^2 0.5(5)(12)^2 KE = 360J
B. PE = mgh = (5)(9.8)(2.6) PE = 127.4J
C. ME = KE + PE = 360 + 127.4 ME = 487.4J
