An ampere (AM-pir), or amp
The acceleration and distance is related to the following expression:
y=v0*t + a*t^2/2 ; v0=0
y=44.1*100/2 = 2205m
hence, the speed will be
v=0 + a*t = 441m/s
from that height it will just be subjected to the gravitational acceleration
0=v_acc^2 -2g*y_free
y_free = v_acc^2/2g = 9922.5m
<span>y_max = y_acc+y_free = 441+9922.5 =10363.5m</span>
Answer:
V = 20.5 m/s
Explanation:
Given,
The mass of the cart, m = 6 Kg
The initial speed of the cart, u = 4 m/s
The acceleration of the cart, a = 0.5 m/s²
The time interval of the cart, t = 30 s
The final velocity of the cart is given by the first equation of motion
v = u + at
= 4 + (0.5 x 30)
= 19 m/s
Hence the final velocity of cart at 30 seconds is, v = 19 m/s
The speed of the cart at the end of 3 seconds
V = 19 + (0.5 x 3)
= 20.5 m/s
Hence, the final velocity of the cart at the end of this 3.0 second interval is, V = 20.5 m/s
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