Using conservation of energy and momentum we get m1*v1=(m1+m2)*v2 so rearranging for v2 and plugging the given values in we get:
(200000kg*1.00m/s)/(21000kg)=.952m/s
Answer:
The Magnifying power of a telescope is 
Explanation:
Radius of curvature R = 5.9 m = 590 cm
focal length of objective
= 
⇒
= 
⇒
= 295 cm
Focal length of eyepiece
= 2.7 cm
Magnifying power of a telescope is given by,



therefore the Magnifying power of a telescope is 
At the given erro in angle, the error in the measurement of sin 90 degrees would be 0.001.
<h3>
Percentage error</h3>
The percentage error of any measurement is obtained from the ratio of the error to the actual measurement.
The error of sin 90 degrees is calculated as follows;
sin 90 = 1
error in measurement = sin(90 - 0.5)
error in measurement = sin(89.5) = 0.999
<h3>Error in sin 90 degrees</h3>
Error in sin 90 degrees = 1 - 0.999
Error in sin 90 degrees = 0.001
Thus, at the given erro in angle, the error in sin 90 degrees would be 0.001.
Learn more about error in measurement here: brainly.com/question/26668346
Answer:
3+7= 7x3= 21➗7=
Explanation:
3+7= 10 7 8 9 10
7x3= 21 7 14 21
21➗7=3 21 14 7
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Answer:
The net emissions rate of sulfur is 1861 lb/hr
Explanation:
Given that:
The power or the power plant = 750 MWe
Since the power plant with a thermal efficiency of 42% (i.e. 0.42) burns 9000 Btu/lb coal, Then the energy released per one lb of the coal can be computed as:

= 3988126.8 J
= 3.99 MJ
Also, The mass of the burned coal per sec can be calculated by dividing the molecular weight of the power plant by the energy released per one lb.
i.e.
The mass of the coal that is burned per sec 
The mass of the coal that is burned per sec = 187.97 lb/s
The mass of sulfur burned 
= 2.067 lb/s
To hour; we have:
= 7444 lb/hr
However, If a scrubber with 75% removal efficiency is utilized,
Then; the net emissions rate of sulfur is (1 - 0.75) × 7444 lb/hr
= 0.25 × 7444 lb/hr
= 1861 lb/hr
Hence, the net emissions rate of sulfur is 1861 lb/hr