The weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
To find the answer, we have to know about the pressure.
<h3>How to find the weight of a column of air?</h3>
where; F is the force, here it is equal to the weight of the air column, and A is the area of cross section.
Thus, we can conclude that, the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
Learn more about the pressure here:
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Answer:
Explanation:
Expression for time period of a pendulum is as follows
T =
l is length of pendulum from centre of bob and g is acceleration due to gravity
Given
Time period T = 1.583
g = 9.846
Substituting the values
1.583 =
l =
l = .6244 m
= 62.44 cm
Length of rod = length of pendulum - radius of bob
= 62.44 - 13.62
= 48.82 cm
= .488 m
Answer:
the work done to lift the bucket = 3491 Joules
Explanation:
Given:
Mass of bucket = 10kg
distance the bucket is lifted = height = 11m
Weight of rope= 0.9kg/m
g= 9.8m/s²
initial mass of water = 33kg
x = height in meters above the ground
Let W = work
Using riemann sum:
the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)
=
Work done in lifting the bucket (W) = force × distance
Force (F) = mass × acceleration due to gravity
Force = 9.8 * 10 = 98N
W done by bucket = 98×11 = 1078 Joules
Work done to lift the rope:
At Height of x meters (0≤x≤11)
Mass of rope = weight of rope × change in distance
= 0.8kg/m × (11-x)m
W done = integral of (mass×g ×distance) with upper 11 and lower limit 0
W done =
Note : upper limit is 11 not 1, problem with math editor
W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0
W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]
=7.84(60.5 -0) = 474.32 Joules
Work done in lifting the water
At Height of x meters (0≤x≤11)
Rate of water leakage = 36kg ÷ 11m = kg/m
Mass of rope = weight of rope × change in distance
= kg/m × (11-x)m = 3.27kg/m × (11-x)m
W done = integral of (mass×g ×distance) with upper 11 and lower limit 0
W done =
Note : upper limit is 11 not 1, problem with math editor
W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0
W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]
= 32.046(60.5 -0) = 1938.783 Joules
the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)
the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103
the work done to lift the bucket = 3491 Joules