Answer:
N = 177843 sheets
Explanation:
We are given;
Mass;m = 0.0035 kg
Pressure; p = 101325 pa = 101325 N/m²
L = 0.279m
W = 0.216m
The weight of N sheets is N(mg)
Where;
m is the mass of one sheet
N is number of sheets
g is the acceleration due to gravity.
The pressure equals weight divided by the area on which the weight presses:
Thus,
p= F/A = Nmg/(L•W)
Therefore, making N the subject;
N = pLW/(mg)
N = 101325 x 0.279 x 0.216/ (0.0035 x 9.81)
N = 177843
Answer:
a) a geostationary satellite is that it is always at the same point with respect to the planet,
b) f = 2.7777 10⁻⁵ Hz
c) d) w = 1.745 10⁻⁴ rad / s
Explanation:
a) The definition of a geostationary satellite is that it is always at the same point with respect to the planet, that is, its period of revolutions is the same as the period of the planet
- T = 10 h (3600 s / 1h) = 3.6 104 s
b) the period the frequency are related
T = 1 / f
f = 1 / T
f = 1 / 3.6 104
f = 2.7777 10⁻⁵ Hz
c) the distance traveled by the satellite in 1 day
The distance traveled is equal to the length of the circumference
d = 2pi (R + r)
d = 2pi (69 911 103 + 120 106)
d = 1193.24 m
d) the angular velocity is the angle traveled between the time used.
.w = 2pi /t
w = 2pi / 3.6 10⁴
w = 1.745 10⁻⁴ rad / s
how fast is
v = w r
v = 1.75 10-4 (69.911 106 + 120 106)
v = 190017 m / s
Answer: True
Explanation:
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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
