Answer:
18.4 m
Explanation:
(a)
The known variables in this problem are:
u = 1.40 m/s is the initial vertical velocity (we take downward direction as positive direction)
t = 1.8 s is the duration of the fall
a = g = 9.8 m/s^2 is the acceleration due to gravity
(b)
The vertical distance covered by the life preserver is given by
If we substitute all the values listed in part (a), we find
Answer:
304.89m
Explanation:
Given
acceleration a = 2.52m/s²
final speed v = 39.2m/s
initial speed = 0m/s (car accelerates from rest)
Using the equation of motion below to get the distance of Doc brown from Marty;
v² = u²+2as
substitute the given parameters
39.2² = 0²+2(2.52)s
1536.64 = 0+5.04s
divide both sides by 5.04
1536.64/5.04 = 5.04s/5.04
rearrange the equation
5.04s/5.04 = 1536.64/5.04
s = 304.89m
Hence He and Marty must stand at 304.89m to allow the car to accelerate from rest to a speed of 39.2 m/s?
Answer:
Explanation:
parallel capacitances add directly
Series capacitances add by reciprocal of sum of reciprocals.
Ceq = [ C ] + [1 / (1/C + 1/C)] + [1 / (1/C + 1/C + 1/C)]
Ceq = [ C ] + [C / 2] + [C / 3]
Ceq = [ 6C/6 ] + [3C / 6] + [2C / 6]
Ceq = 11C/6
