To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.
Newton's second law is defined as

Where,
m = mass
a = acceleration
From this equation we can figure the acceleration out, then



From the cinematic equations of motion we know that

Where,
Final velocity
Initial velocity
a = acceleration
x = displacement
There is not Final velocity and the acceleration is equal to the gravity, then





From the equation of motion where acceleration is equal to the velocity in function of time we have




Therefore the time required is 0.0705s
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The value is 
Explanation:
From the question we are told that
The length of the stretcher is 
The weight of the stretcher is 
The weight for Wayne is 
The distance of center of gravity for Wayne from Chris is 
Generally taking moment about the first end where Chris is
=> upward moment
Here
is the force applied by Jamie
Generally taking moment about the second end where Jamie is
=> downward moment
Generally at equilibrium , the upward moment is equal to the downward moment

=> 
=> 
Answer:
Explanation:
If Bradley examination was done and interpreted in the same facility, the radiologist code is used example- procedure code 72100- Radiologic examination, spine, lumbosacral, 2 or 3 views is reported.
if the X-ray was taken by Dr X but Dr X does not read or interpret the image but forward it to the radiologist for initial report, then a 26- modifier is used. E.g A reports by the technologist would be, procedure code 72050-Radiologic examination, spine, cervical, 2 or 3
views or 72050- TC in certain situations and the consulting radiologist would report 72050-26.
if Bradley’s x-ray were sent to an independent radiologist for interpretation, then the procedure code 76140 is used in reporting.
Answer:
The acceleration of the cheetahs is 10.1 m/s²
Explanation:
Hi there!
The equation of velocity of an object moving along a straight line with constant acceleration is the following:
v = v0 + a · t
Where:
v = velocity of the object at time t.
v0 = initial velocity.
a = acceleration.
t = time
We know that at t = 2.22 s, v = 50.0 mi/h. The initial velocity, v0, is zero.
Let's convert mi/h into m/s:
50.0 mi/h · (1609.3 m / 1 mi) · (1 h / 3600 s) = 22.4 m/s
Then, using the equation:
v = v0 + a · t
22.4 m/s = 0 m/s + a · 2.22 s
Solving for a:
22.4 m/s / 2.22 s = a
a = 10.1 m/s²
The acceleration of the cheetahs is 10.1 m/s²
Answer:
The maximum speed of the car at the bottom of that drop is 26.34 m/s.
Explanation:
Given that,
The maximum vertical distance covered by the roller coaster, h = 35.4 m
We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :



v = 26.34 m/s
So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.