Answer:
a. 2.0secs
b. 20.4m
c. 4.0secs
d. 141.2m
e. 40m/s, ∅= -30°
Explanation:
The following Data are giving
Initial speed U=40m/s
angle of elevation,∅=30°
a. the expression for the time to attain the maximum height is expressed as
where g is the acceleration due to gravity, and the value is 9.81m/s if we substitute values we arrive at
b. the expression for the maximum height is expressed as
c. The time to hit the ground is the total time of flight which is twice the time to reach the maximum height ,
Hence T=2t
T=2*2.0
T=4.0secs
d. The range of the projectile is expressed as
e. The landing speed is the same as the initial projected speed but in opposite direction
Hence the landing speed is 40m/s at angle of -30°
Answer:
Distance, Brighter
Explanation:
Think of a flash light, one close and one far which is brighter and which is dimmer.
Hope this is what your teacher put as the answer!
Answer:
a) 12.74 V
b) Two pairs of diode will work only half of the cycle
c) 8.11 V
d) 8.11 mA
Explanation:
The voltage after the transformer is relationated with the transformer relationshinp:
the peak voltage before the bridge rectifier is given by:
The diodes drop 0.7v, when we use a bridge rectifier only two diodes are working when the signal is positive and the other two when it's negative, so the peak voltage of the load is:
As we said before only two diodes will work at a time, because the signal is half positive and half negative,so two of them will work only half of the cycle.
The averague voltage on a full wave rectifier is given by:
Using Ohm's law:
Given:
h = 600 m, the height of descent
t = 5 min = 5*60 = 300 s, the time of descent.
Let a = the acceleration of descent., m/s².
Let u = initial velocity of descent, m/s.
Let t = time of descent, s.
The final velocity is v = 0 m/s because the helicopter comes to rest on the ground.
Note that u, v, and a are measured as positive upward.
Then
u + at = v
(u m/s) + (a m/s²)*(t s) = 0
u = - at
u = - 300a (1)
Also,
u*t + (1/2)at² = -h
(um/s)*(t s) + (1/2)(a m/s²)*(t s)² = 600
ut + (1/2)at² = 600 (2)
From (1), obtain
-300a +(1/2)(a)(90000) = -600
44700a = -600
a = - 1.3423 x 10⁻² m/s²
From (1), obtain
u = - 300*(-1.3423 x 10⁻²) = 4.03 m/s
Answer:
The acceleration is 0.0134 m/s² downward.
The initial velocity is 4.0 m/s upward.
