Given
initial position = Xi= 19.9m
Final position Xf = 5.4m
Average velocity= Va = -0.418m/s
it shows displacement is reverse.
To find t=?
As Va = (Xf- Xi) / t
t = (Xf-Xi) / ( Va)
t = ( 5.4-19.9) / (-0.418)
t = (-14.5 ) / (-0.418) (-ve sign cancel out at numerator and denominator)
t =34.69 s
okay this is kinda easy
<u>What is the gravitational field strength on the moon?</u>
The Moon has a gravitational field strength of 1.6 N/kg.
Answer:
Explanation:
Remark
The only thing that might trip you up is what to do with the angle. The vertical component of the 15 degrees does no work against anything. So the 15 degrees limits the horizontal force.
The formula is
Work = F * d * cos(15)
The givens are
F = 2000 N
d = 30 m
Cos(15) = 0.9659
Solution
Work = 2000 * 30 * cos(15)
Work = 57,955
Rounded to two places would be 5.8 * 10^4
C
Answer:
the magnitude of Vpg = 493.711 km/h
Explanation:
given data
speed Vpg = 560 km/h
speed Vwg = 80 km/h
solution
we get here magnitude of the plane velocity w.r.t. ground is
we know that the Vpg = Vpw + Vwg .....................1
writing the component of the velocity that is
Vpw = (0 km/h î - 560 km/h j )
Vwg = (80 cos 45 km/h î + 80 sin 45 km/h j)
adding these
Vpg = (0+80 cos 45 km/h ) î + ( -560 + 80 sin 45 km/h j)i
Vpg = (42.025 ) î (-491.92 km/h)j
now we take magnitude
the magnitude of Vpg = 
the magnitude of Vpg = 493.711 km/h
