Answer:
Explanation:
A ) angular velocity ω = 2π / T
= 2 x 3.14 / 60
= .10467 rad / s
linear velocity v = ω R
= .10467 x 50
= 5.23 m / s
centripetal force = m v² / R
= mg v² / gR
= 834 x 5.23² / 9.8 x 50
= 46.55 N
B )
apparent weight
= mg - centripetal force
= 834 - 46.55
= 787.45 N
C ) apparent weight
= mg + centripetal force
= 834 + 46.55
= 880.55 N.
D )
For apparent weight to be zero
centripetal force = mg
mg = mv² / R
v² = gR
= 9.8 x 50
= 490
v = 22.13 m /s
time period of revolution
= 2π R /v
2 x 3.14 x 50 / 22.13
= 14.19 s
Answer:
<em>The object-Earth system is open</em>

Explanation:
<u>Accelerated Motion
</u>
When an object is released in free air (with no other forces than the gravity), it describes a free-fall motion and the formulas include the acceleration of gravity as part of the calculations. But when there is another external force, then the acceleration is not the gravity, but the result of the net force exerted on the mass of the object.
By definition, an open system includes the exchange of energy from and to the surroundings, that is why all systems surrounding our planet are considered as open systems. In our case, the object is interacting with the planet's gravity and there is some other external force, which will be computed later. The object-Earth system is open.
If the object starts from rest, its initial speed is zero, and

where a is the acceleration and t is the time. The distance traveled is given by
:

From the two above equations, we find that:

Solving for a



It means the net force is

The object's weight is

This means there is some external force acting upwards delaying the object's fall of a magnitude of

First we have to convert:
75 km / h ( * 3.6 ) = 270 m/s
18 m/s = 64.8 m/s
d = v i · t + 1/2 a · t²
v = v i + a t
-----------------------------------
4000 = 64.8 · t + 1/2 a · t²
270 = 64.8 + a t
a = ( 270 - 64.8 ) / t
a = 205.2 / t
4000 = 64.8 t + 1/2 · ( 205.2 / t ) · t²
4000 = 64.8 t + 102.6 t
4000 = 167.4 t
t = 4000 : 167.4 = 23.89 s
a = 205.2 : 23.89 = 8.58 m/s²
Answer: His average acceleration was 8.58 m/s²
Answer - An object is in motion when its distance from another object is changing. ... A reference point is a place or object used for comparison to determine if something is in motion. An object is in motion if it changes position relative to a reference point. You assume that the reference point is stationary, or not moving.