Answer:
d. low earth orbit (LEO)
Explanation:
This type of satellites form a constellation deployed as a series of “necklaces” in such a way that at any time, at least one satellite is visible by a receiver antenna, compensating the movement due to the earth rotation.
Opposite to that, a geostationary satellite is at an altitude that makes it like a fixed point over the Earth´s equator, rotating synchronously with the Earth, so it is always visible in a given area.
Answer:
the crown is false densty= 12556kg/m^3[/tex]
Explanation:
Hello! The first step to solve this problem is to find the mass of the crown, this is found using the weight of the crown in the air by means of the equation for the weight.
W=mg
W=weight(N)=31.4N
M=Mass
g=gravity=9.81m/S^2
solving for M
m=W/g

The second step is find the volume of crown remembering that when an object is weighed in the water the result is the subtraction between the weight of the object and the buoyant force of the water which is the product of the volume of the crown by gravity by density of water

Where
F=weight in water=28.9N
m=mass of crown=3.2kg
g=gravity=9.81m/S^2
α=density of water=1000kg/m^3
V= crown´s volume
solving for V

finally, we remember that the density is equal to the index between mass and volume

To determine the density of the crown without using the weight in the water and with a bucket we can use the following steps.
1.weigh the crown in the air and find the mass
2. put water in a cylindrical bucket and measure its height with a ruler.
3. Put the crown in the bucket and measure the new water level with a ruler.
4. Subtract the heights, and find the volume of a cylinder knowing the difference in heights and the diameter of the bucket, in order to determine the volume of the crown.
5. find density by dividing mass by volume
Answer:
471 days
Explanation:
Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons
As,
1 gallon = 0.133 cubic feet (cf)
Therefore,
Capacity of Carvins Cove water reservoir in cf = 3.2 x 10˄9 x 0.133
= 4.28 x 10˄8
Applying Mass balance i.e
Accumulation = Mass In - Mass out (Eq. 01)
Here
Mass In = 0.5 cfs
Mass out = 11 cfs
Putting values in (Eq. 01)
Accumulation = 0.5 - 11
= - 10.5 cfs
Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.
Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600
= 37,800
Converting depletion of reservoir in cubic feet per day = 37, 800 x 24
= 907,200
i.e. 907,200 cubic feet volume is being depleted in days = 1 day
1 cubic feet volume is being depleted in days = 1/907,200 day
4.28 x 10˄8 cubic feet volume will deplete in days = (4.28 x 10˄8) x 1/907,200
= 471 Days.
Hence in case of continuous drought reservoir will last for 471 days before dry-up.
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average speed (in km/h) of a car stuck in traffic that drives 12 kilometers in 2 hours.
Answer:
See the attached picture for answer.
Explanation:
See the attached picture for explanation.