How fast would the International Space Station (ISS) have to travel to maintain a circular orbit a distance of 1400 km above the
1 answer:
Answer:
The International Space Station move at 7.22 km/s.
Explanation:
Orbital speed of satellite is given by , where G is gravitational constant, M is mass of Earth and r is the distance to satellite from centre of Earth.
r = R + h = 6350 + 1400 = 7750 km = 7.75 x 10⁶ m
G = 6.673 x 10⁻¹¹ Nm²/kg²
M = 5.98 x 10²⁴ kg
Substituting
The International Space Station move at 7.22 km/s.
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Answer:
t = 4 s
Explanation:
As we know that the particle A starts from Rest with constant acceleration
So the distance moved by the particle in given time "t"
Now we know that B moves with constant speed so in the same time B will move to another distance
now we know that B is already 349 cm down the track
so if A and B will meet after time "t"
then in that case
on solving above kinematics equation we have
Answer:
The 5-number summary is
1. Median = 0.93 W/kg
2. Lower quartile = 0.69 W/kg
3. Upper quartile = 1.16 W/kg
4. Minimum value = 0.54 W/kg
5. Maximum value = 1.42 W/kg
Explanation:
We are given the measured radiation absorption rates (in W/kg) corresponding to 11 cell phones.
1.16 0.85 0.69 0.75 0.95 0.93 1.18 1.17 1.42 0.54 0.57
What is 5-number summary?
A 5-number summary refers to a box plot that basically shows 5 statistical characteristics of a data set.
These statistical characteristics are:
1. Median
2. Lower quartile
3. Upper quartile
4. Minimum value
5. Maximum value
1. Median:
Arrange the data in ascending order
0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42
(n+1)/2 gives the median value of the data set.
(11 + 1)/2 = 6th position
Therefore, 0.93 W/kg is the median of the data set.
2. Lower quartile:
Divide the data set into two equal halfs (include median in both if n = odd)
Lower half = 0.54 0.57 0.69 0.75 0.85 0.93
Upper half = 0.93 0.95 1.16 1.17 1.18 1.42
The lower quartile is the median of the lower half of the data set.
Lower half = 0.54 0.57 0.69 0.75 0.85 0.93
The median is 6/2 = 3rd position
Therefore, the lower quartile of the data set is 0.69 W/kg
3. Upper quartile:
Divide the data set into two equal halfs (include median in both if n = odd)
Lower half = 0.54 0.57 0.69 0.75 0.85 0.93
Upper half = 0.93 0.95 1.16 1.17 1.18 1.42
The upper quartile is the median of the lower half of the data set.
Upper half = 0.93 0.95 1.16 1.17 1.18 1.42
The median is 6/2 = 3rd position
Therefore, the upper quartile of the data set is 1.16 W/kg
4. Minimum value:
The minimum value is the least value in the data set.
0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42
Therefore, the minimum value of the data set is 0.54 W/kg
5. Maximum value
The maximum value is the least value in the data set.
0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42
Therefore, the maximum value of the data set is 1.42 W/kg
The box plot is illustrated in the attached diagram.
Answer:
-414.96 N
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
The force the ground exerts on the parachutist is -414.96 N
If the distance is shorter than 0.75 m then the acceleration will increase causing the force to increase