Answer:
The correct answer to the following question will be "41.87 m".
Explanation:
The given values are:
The speed of trooper = 
The velocity of red car = 
Now,
A red car goes as far as possible until the speed or velocity of the troops is the same as that of of the red car at
(∵
)

then,
The distance covered by trooper,


The distance covered by red car,
= 
= 
Maximum distance = 
=
about 5 watts (5W) of power
Mnanan anaw anwkkww wjakana akkqna akaoaman akamakq
Answer:
F = 32.28 N
Explanation:
For this exercise we must use the rotational equilibrium relation
Σ τ = 0
In the initial configuration it is in equilibrium, for which all the torque and forces are compensated. By the time the payment lands on the bar, we assume that the counter-clockwise turns are positive.
W_bird L / 2 - F_left 0.595 - F_right 0.595 = 0
we assume that the magnitude of the forces applied by the hands is the same
F_left = F_right = F
W_bird L / 2 - 2 F 0.595 = 0
F =
we calculate
F = 0.560 9.8 14.0 /2.38
F = 32.28 N
Answer:
a) 725.5 m
b) 630 m
Explanation:
Given data:
acceleration of Helicopter = 7.0 m/s^2
time spent upwards by helicopter = 11.0 seconds
a) Determine the maximum height above ground reached by the helicopter
h1 = at^2 /2
= 7 * 11^2 / 2
= ( 7 * 121 ) / 2 = 423.5 m
also v = a*t = 7 * 11 = 77 m/s
also we calculate h2
h2 = v^2 / 2g
= (77^2) / 2 * 9.81
= 302 m
therefore the maximum height = 302 + 423.5 = 725.5 m
b) Given that ; power deploys a jet pack strapped on his back at 7.0 s and with a downward acceleration of ; 1.0 m/s^2
<u>Determine distance Power reaches before helicopter crashes </u>
s = ut + 1/2 at^2
h.gt^2 - 77t - 423.5 m = 0
h.gt^2 - 77t = 423.5
t = 17. 66 secs
Yf = 423.5 + 77 *7 - 4.9 *7
Yf = 928.2
Vf = u + at
= 77 - 9.8*7 = 8.4 m/secs
t' = 17.66 - 7 = 10.66 secs
hence
Yf = 725.5 - 8.4 * 10.66 + 1/2 * -1 * 10.66
= 630 meters