Answer:
A. 2.30 x 10^2 kil
Explanation:
3/2 = 1.5 and 0.72/2 = 0.36
1.93 + 0.36 = 2.29
2.29 = about 2.30
consider east-west direction along x-axis and north-south direction along y-axis
from the diagram
A = Ax i + By j = - (0.40 Sin60) i + (0.40 Cos60) j = - 0.35 i + 0.20 j
B = Bx i + By j = - 0.50 i + 0 j
Net displacement is given as
D = A + B
D = (- 0.35 i + 0.20 j ) + (- 0.50 i + 0 j )
D = - 0.85 i + 0.2 j
magnitude of displacement is given as
|D| = sqrt((- 0.85)² + (0.2)²)
|D| = 0.87 km
direction of displacement is given as
θ = tan⁻¹(0.2/0.85)
θ = 13.5 deg north of west
Closest to the sun f thats a choice or closest to the atmoshpere
Q: A rock is thrown off of a 100 foot cliff with an upward velocity of 45 m/s. As a result its height after t seconds is given by the formula:
h(t)=100+45t−4.9t2
(a)
What is its height after 3 seconds?
(b)What is its velocity after 3 seconds?
Answer:
(a) 190.9 m.
(b) 15.6 m/s upward
Explanation:
Given:
h(t) = 100 + 45t - 4.9t²
The height after 3 seconds,
t = 3 s
Substitute the value of t in to the equation above.
h(3) = 100+45(3)-4.9(3)²
h(3) = 100+135-44.1
h(3) = 190.9 m
Therefore the height after 3 seconds = 190.9 m.
(b) Velocity after 3 seconds
The velocity is obtained by differentiating h(t) with respect to time
v = dh(t)/dt
dh(t)/dt = 45-9.8t
v = 45 - 9.8t ......................................... Equation 1
t = 3 s.
Substitute the value of t into the equation above,
v = 45 - 9.8(3)
v = 45- 29.4
v = 15.6 m/s
Thus the velocity after 3 seconds = 15.6 m/s upward