Answer:
The rocket has to be launched 8 m from the hoop
Explanation:
Let's analyze this problem, the rocket is on a car that moves horizontally, so the rocket also has the same speed as the car; The initial horizontal rocket speed is (v₀ₓ = 3.0 m/s).
On the other hand, when starting the engines we have a vertical force, which creates an acceleration in the vertical axis, let's use Newton's second law to find this vertical acceleration
F -W = m a
a = (F-mg) / m
a = F/m -g
a = 7.0/0.500 - 9.8
a = 4.2 m/s²
We see that we have a positive acceleration and that is what we are going to use in the parabolic motion equations
Let's look for the time it takes for the rocket to reach the height (y = 15m) of the hoop, when the rocket fires its initial vertical velocity is zero (I'm going = 0)
y = t + ½ a t²
y = 0 + ½ a t²
t = √ 2y/a
t = √( 2 15 / 4.2)
t = 2.67 s
This time is also the one that takes in the horizontal movement, let's calculate how far it travels
x = v₀ₓ t
x = 3 2.67
x = 8 m
The rocket has to be launched 8 m from the hoop
Answer:
Check the explanation
Explanation:
This is the step by step explanation to the above question:
= v * (83.1 * (v-4.3) - 80.7 ( v+4.3))/ [83.1 *(v - 4.3) + 80.7*(v + 4.3)]
v = 344 m/s
vi = 344 * ( 83.1* (344-4.3) - 80.7*(344+4.3) ) / (83.1 *(344 - 4.3) + 80.7*(344 + 4.3))
= 0.74 m/s
Answer:
<em>h = 20 m</em>
Explanation:
<u>Gravitational Potential Energy</u>
Gravitational potential energy (GPE) is the energy stored in an object due to its vertical position or height in a gravitational field.
It can be calculated with the equation:
U=m.g.h
Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or .
The weight of an object of mass m is:
W = m.g
Thus, the GPE is:
U=W.h
Solving for h:
The weight of the owl is W=40 N and its GPE is U=800 J.
h = 20 m
With constant angular acceleration , the disk achieves an angular velocity
at time
according to
and angular displacement according to
a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of
b. Under constant acceleration, the average angular velocity is equivalent to
where and
are the final and initial angular velocities, respectively. Then
c. After 1.00 s, the disk has instantaneous angular velocity
d. During the next 1.00 s, the disk will start moving with the angular velocity equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle
according to
which would be equal to