On sources it says it would just be the super giant star
<span>49N is the force needed to give a .25 kg arrow an acceleration of 196m/s2. F =ma ⇒ =( 0.25kg)(196m/s2) = 49N if the arrow is shot horizontally where the applied force is entirely in the x-direction.</span>
Answer:
The length of her shadow is changing at the rate -2 m/s
Explanation:
Let the height oh the street light, h = 22 ft
Let the height of the woman, w = 5.5 ft
Horizontal distance to the street light = l
length of shadow = x
h/w = (l + x)/x
22/5.5 = (l + x)/x
4x = l + x
3x = l
x = 1/3 l
taking the derivative with respect to t of both sides
dx/dt = 1/3 dl/dt
dl/dt = -6 ft/sec ( since the woman is walking towards the street light, the value of l is decreasing with time)
dx/dt = 1/3 * (-6)
dx/dt = -2 m/s
Answer:
(A) 1.43secs
(B) -2.50m/s^2
Explanation:
A commuter backs her car out of her garage with an acceleration of 1.40m/s^2
(A) When the speed is 2.00m/s then, the time can be calculated as follows
t= Vf-Vo/a
The values given are a= 1.40m/s^2 , Vf= 2.00m/s, Vo= 0
= 2.00-0/1.40
= 2.00/1.40
= 1.43secs
(B) The deceleration when the time is 0.800secs can be calculated as follows
a= Vf-Vo/t
= 0-2.00/0.800
= -2.00/0.800
= -2.50m/s^2
