Answer:
The horizontal component of its velocity remains constant and the vertical component of its acceleration is equal to -g.
Explanation:
This is because, the projectile has both vertical and horizontal components of velocity. But, its vertical component of velocity changes as the object moves whereas, its horizontal component of velocity remains constant.
Also, the projectile has only vertical component of acceleration and no horizontal component of acceleration since, its horizontal component of velocity remains constant. Thus, no change in the horizontal component of velocity.
The vertical component of acceleration is equal to -g since, the weight is the only vertical force acting on it.
So, <u>the horizontal component of its velocity remains constant and the vertical component of its acceleration is equal to -g.</u>
Answer:
Using the range formula R = v^2 sin 2 theta / g
or v^2 = R * g / sin 86.4
v^2 = 3.14 m * 9.81 m/s2 / .998
v^2 = 30.9 m^2 / s^2
v = 5.56 m/s
This hasn't really proved the question - this would give
vy = 5.56 * sin 43.2 = 3.81 m/s
vx = 5.56 * cos 43.2. = 4.05 m/s
t = 1.57 / 4.05 = .387 sec to reach the waterfall
h = 3.81 * .387 - 4.9 (.387)^2 = .74 m well above the height of the falls
There seems another way to do this
vy / vx = tan 43.2 vy = .939 vx
h = vy t - 1/2 g t^2 and t = 1.57 / vx
h = 1.57 tan 43.2 - 4.9 (1.57 / vx)^2
Solving for vx I get vx = 3.26 m/s vy = 3.06 m/s v = 4.47 m/s
Answer: 0.313 rad/s
Explanation:
The equation that relates the velocity 
and the angular velocity 
in the uniform circular motion is:
(1)
Where 
is the radius of the space station (with a diaeter of 200m) that describes the uniform circular motion.
Isolating from (1):
(2)
On the other hand, we are told the “artificial gravity” produced by the cetripetal acceleration 
is 
, and is given by the following equation:
(3)
Isolating :
(4)
(5)
Substitutinng (5) in (2):
(6)
This is the angular velocity that would produce an “artificial gravity” of 9 .
Answer:
A) 588 pounds
Explanation:
According to the given conditions, we assume the beam to be simply supported at the ends carrying a uniformly distributed load of 125 pounds per feet and a point load of 600 pounds acting at 5 feet from the right support.
Referring the schematic:
<u>Moment about any point will be zero in equilibrium condition.
</u>
∴Take moment about point L
Answer:
Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
Explanation:
