From research I found that the <span>fossil fuel resources are sometimes referred to as "buried sunshine" because they make the earth warm. They are also found deep in the earth and the energy stored in them was obtained from the sun long ago, and that's because they also produce energy the same way the sun does.
Hope this is the answer you were looking for and that it helps. :)</span>
Answer:
b = 5.164 m is the minimum thickness of the dam
Explanation:
Given
SG = 5
h = 20 m
b = ?
w = 1 m
γw = 9800 N/m³
We can get the forces as follows
Fp = γw*(h/2)*(h*w)
⇒ Fp = (9800 N/m³)*(20 m/2)*(20 m*1 m) = 1.96*10⁶N
W = (SG*γw)*(h*b*w)
⇒ W = (5*9800 N/m³)*(20 m*b*1 m) = (9.8*10⁵N/m)*b
Then, we apply
∑M₀ = 0 (counterclockwise)
- Fp*(h/3) + W*(b/2) = 0
⇒ - 1.96*10⁶N*(20 m/3) + (9.8*10⁵N/m)*b*(b/2) = 0
- 13.066*10⁶N-m + (4.9*10⁵N/m)*b² = 0
⇒ b = 5.164 m is the minimum thickness of the dam
Answer:
1/ decreases
2/ yes;
In general there is no relationship, and any combinations of frequencies and amplitudes are allowed.
Explanation:
Explanation for 1/ - please see the picture attached below
Answer:
hello the diagram relating to this question is attached below
a) angular accelerations : B1 = 180 rad/sec, B2 = 1080 rad/sec
b) Force exerted on B2 at P = 39.2 N
Explanation:
Given data:
Co = 150 N-m ,
<u>a) Determine the angular accelerations of B1 and B2 when couple is applied</u>
at point P ; Co = I* ∝B2'
150 = ( (2*0.5^2) / 3 ) * ∝B2
∴ ∝B2' = 900 rad/sec
hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec
at point 0 ; Co = Inet * ∝B1
150 = [ (2*0.5^2) / 3 + (2*0.5^2) / 3 + (2*0.5^2) ] * ∝B1
∴ ∝B1 = 180 rad/sec
hence angular acceleration of B1 = 180 rad/sec
<u>b) Determine the force exerted on B2 at P</u>
T2 = mB1g + T1 -------- ( 1 )
where ; T1 = mB2g ( at point p )
= 2 * 9.81 = 19.6 N
back to equation 1
T2 = (2 * 9.8 ) + 19.6 = 39.2 N
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Answer:
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Explanation: