S = ut + 0.5at^2
<span>10 = 0 + 0.5(9.81)t^2 {and if g = 10 then t^2 = 2 so t ~1.414} </span>
<span>t^2 ~ 2.04 </span>
<span>t ~ 1.43 seconds</span>
The correct answer is A. B. C. Hope this helps!!
GIVE vo0Mac0ov Brainliest ... He's smart
Find Displacement and Distance
displacement ...
north is 700+400+100 =1200m n
south=1200m
1200-1200=0
east is 300+300=600m
west is 600m
600-600=0
back at dtart. displ zero
distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m = 3600m
Answer:
Explanation:
We shall represent speed in vector form
First speed
v₁ = 1.5 cos 14 i + 1.5 sin 14 j
= 1.455 i + 0.363 j
v₂ = 4.4 cos 33 i + 4.4 sin 33 j
= 3.69 i + 2.39 j
v₂ - v₁
3.69 i + 2.39 j - 1.455 i - 0.363 j
= 2.235 i + 2.027 j
acceleration
= v₂ - v₁ / time
= ( 2.235 i + 2.027 j ) / 23
= .097 i + .088 j
force = mass x acceleration
= 398 x ( .097 i + .088 j )
= 38.6 i + 35.02 j
Magnitude of force F
F² = 38.6² + 35.02²
F = 52.11 N
Tan θ = 35.02 / 38.6
θ = 42° north of east.