At 8:00 pm, the velocity of the storm is 55 mi northeast. Assuming that the direction is exactly northeast, the angle is 45°
At 11:00 pm, the velocity is 75 mi north. The angle is 90°
In vector form
55 ∠ 45°
and
75 ∠ 90°
The magnitude and direction of the average velocity is
(55 ∠ 45° + 75 ∠ 90° ) / 3
Answer:
a) 1.75s b) 17.2 m/s (down)
Explanation:
d1= 15m d2= 0m (because it hits ground)
a= -9.81 m/s^2 t=???
Equation
the triangle means change in so d2-d1
Δd= v1 * t + 1/2 * a * t^2
0m-15m= v1*t + 1/2 a t^2
-15 m= 0m/s*t (goes away) + 1/2* a *t^2
-15mx2= t^2
-15mx2/a= t^2
Square root (-30/-9.81m/s^2)
t=1.75 s
b) now v2!!
Im going to use v2= v1 + a*t
v2= 0m/s + -9.81 x 1.75s
v2 = -17.2 m/s or you can say 17.2 m/s down!!!
A wave that is oscillation of matter.. such as a water ripples
Answer:
0 m/s
Explanation:
velocity= change in displacement/ time
at rest, the ball does not travel any distance
0/ t
=0
