Answer:
Explanation:
Since the front and back of the rocket simultaneously line up with forward and backward end of the platform respectively .
Then length of the platform = length of the train rocket .
A )
Time to cross a particular point on the platform
= length of rocket train / .96 x 3 x 10⁸
= 90 / .96 x 3 x 10⁸
= 31.25 x 10⁻⁸ s
B) Rest length of the rocket = length of platform = 90 m
C ) length of platform as viewed by moving observer =

= 
= 321 m
D ) For the observer on platform time taken = 31.25 x 10⁻⁸ s
for the observer in the rocket , time will be dilated so time recorded by observer in motion ,
8.75 x 10⁻⁸ s .
The parcel will undergo projectile motion, which means that it will have motion in both the horizontal and vertical direction.
First, we determine how long the parcel will fall using:
s = ut + 1/2 at²
where s will be the height, u is the initial vertical velocity of the parcel (0), t is the time of fall and a is the acceleration due to gravity.
5.5 = (0)(t) + 1/2 (9.81)(t)²
t = 1.06 seconds
Now, we may use this time to determine the horizontal distance covered by the parcel by using:
distance = velocity * time
The horizontal velocity of the parcel will be equal to the horizontal velocity of the cruise liner.
Distance = 10 * 1.06
Distance = 10.6 meters
The boat should be 10.6 meters away horizontally from the point of release.
The hoop is attached.
Consider that the friction force is given by:
F = μ·N
= μ·m·g·cosθ
We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·<span>g·sinθ
Therefore:
</span>-(1/2)·m·g·sinθ + m·g·sinθ = <span>μ·m·g·cosθ
</span>(1/2)·m·g·sinθ = <span>μ·m·g·cosθ
</span>μ = (1/2)·m·g·sinθ / <span>m·g·cosθ
= </span>(1/2)·tanθ
Now, solve for θ:
θ = tan⁻¹(2·μ)
= tan⁻¹(2·0.9)
= 61°
Therefore, the maximum angle <span>you could ride down without worrying about skidding is
61°.</span>
Change in velocity = d(v)
d(v) = v2 - v1 where v1 = initial speed, v2 = final speed
v1 = 28.0 m/s to the right
v2 = 0.00 m/s
d(v) = (0 - 28)m/s = -28 m/s to the right
Change in time = d(t)
d(t) = t2 - t1 where t1 = initial elapsed time, t2 = final elapsed time
t1 = 0.00 s
t2 = 5.00 s
d(t) = (5.00 - 0.00)s = 5.00s
Average acceleration = d(v) / d(t)
(-28.0 m/s) / (5.00 s)
(-28.0 m)/s * 1 / (5.00 s) = -5.60 m/s² to the right
Mechanical energy is the sum of kinetic energy and potential energy, or E=Ek+Ep. So Ek=(1/2)*m*v² where m is the mass of the object and v is it's velocity. Mp=m*g*h where m is the mass, g=9.81 m/s² and h is the height of the object. So after we input the numbers the total mechanical energy is
E=(1/2)*2.5*(4.5²) + 2.5*9.81*18 = 25.3125 J + 441.45 J = 466.7625 J. The correct answer is E= 466 J.