Answer:
The displacement of the train in this time period is 2,616.86 m.
Explanation:
A Uniformly Varied Rectilinear Motion is Rectilinear because the mobile moves in a straight line, Uniformly because of there is a magnitude that remains constant (in this case the acceleration) and Varied because the speed varies, the final speed being different from the initial one.
In other words, a motion is uniformly varied rectilinear when the trajectory of the mobile is a straight line and its speed varies the same amount in each unit of time (the speed is constant and the acceleration is variable).
An independent equation of useful time in this type of movement is:
<em>Expression A</em>
where:
- vf = final velocity
- vi = initial velocity
- a = acceleration
- d = distance
The equation of velocity as a function of time in this type of movement is:
vf=vi + a*t
So the velocity can be calculated as: 
In this case:
- vf=42.4 m/s
- vi=27.5 m/s
- t=75 s
Replacing in the definition of acceleration: 
a=0.199 m/s²
Now, replacing in expression A:

Solving:

d= 2,616.86 m
<u><em>The displacement of the train in this time period is 2,616.86 m.</em></u>
Explanation:
The width of the central maximum is given by
W = 2 λ L / a
where W is the width of the central maximum
λ is the wavelength of the light used.
L is the distance between the aperture and screen
a is the width of the slit or aperture
So we can see that if any one quantity is varied by keeping others constant in the above formula , there would be a change in width of central maximum.
Answer:
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Explanation:
Answer:
390 J
Explanation:
m = 3 kg
u = 16 i + 2 j
(a) Magnitude of velocity =
= 16.1245 m/s
KEi = 1/2 m v^2 = 0.5 x 3 x 16.1245 = 390 J
(b) v = 18 i + 14 j
Magnitude of velocity =
= 22.804 m/s
KEf = 1/2 m v^2 = 0.5 x 3 x 22.804 = 780 J
According to the work energy theorem
Work done = change in KE = KEf - KEi = 780 - 390 = 390 J
Explanation:
(a) Net force acting on the block is as follows.

or, ma = -mg Sin (\theta)[/tex]
a = 
= 
= -3.35 
According to the kinematic equation of motion,

Distance traveled by the block before stopping is as follows.
s = 
= 
= 21.5 m
According to the kinematic equation of motion,
v =
0 = 
= 7.16 sec
Therefore, before coming to rest the surface of the plane will slide the box till 7.16 sec.
(b) When the block is moving down the inline then net force acting on the block is as follows.

ma = 
a = 
= 
= 3.35 
Kinematics equation of the motion is as follows.
s = 
21.5 m = 
= 
= 3.58 sec
Hence, total time taken by the block to return to its starting position is as follows.
t = 
= 7.16 sec + 3.58 sec
= 10.7 sec
Thus, we can conclude that 10.7 sec time it take to return to its starting position.