for all of them its base times height
v=bh
1) (8*6/2)*9.5=228
2) (4.5/2)^2 *8= 40.5
3) (3.3* 8.3/2) *5.4= 74.0
4) ((3.2+4.8)/2)* 2.7 *4.4= 47.5
For this case we have the following equation:
S = 0.159 + 0.118logD
Substituting S = 0.2 we have:
0.2 = 0.159 + 0.118logD
From here, we clear the value of D.
We have then:
0.2 - 0.159 = 0.118logD
logD = (0.2 - 0.159) / (0.118)
logD = 0.347457627
10 ^ (logD) = 10 ^ 0.347457627
D = 10 ^ (0.347457627)
D = 2.23 mm
Answer:
The average diameter, D, of the sand particles for a beach is:
D = 2.23 mm
Answer:
2
Step-by-step explanation:
sin²a + cos²a + sec²a - tan²a
sin²a + cos²a = 1
1 + tan²a = sec²a
So, sec²a - tan²a = 1
sin²a + cos²a + sec²a - tan²a
= 1 + 1 = 2
In this problem, volume is computed:
volume = ∫(flow rate)·dt
.. = ∫[0, t] (3/10)*sin(2πt/7)·dt
= (3 / 10)*(7 /(2 π))*(-cos (2 π t/7)+1)
= (-21 / (20 π))*(cos (2 π t /7) - 1)
Volume is equal to 21 sin^2 (pit/7) / 10pi is the inhaled air in the lungs at time t
