Answer:
Normal stress = 66/62.84 = 1.05kips/in²
shearing stress = T/2 = 0.952/2 = 0.476 kips/in²
Explanation:
A steel pipe of 12-in. outer diameter d₂ =12in d₁= 12 -4in = 8in
4 -in.-thick
angle of 25°
Axial force P = 66 kip axial force
determine the normal and shearing stresses
Normal stress б = force/area = P/A
= 66/ (П* (d₂²-d₁²)/4
=66/ (3.142* (12²-8²)/4
= 66/62.84 = 1.05kips/in²
Tangential stress T = force* cos ∅/area = P/A
= 66* cos 25/ (П* (d₂²-d₁²)/4
=59.82/ (3.142* (12²-8²)/4
= 59.82/62.84 = 0.952kips/in²
shearing stress = tangential stress /2
= T/2 = 0.952/2 = 0.476 kips/in²
Answer:
20 pig callers
Explanation:
Given that:
A pig caller produced intensity level of a sound = 107 dB
To find how many pig callers required to generate an intensity level of 120 dB;
we have:
120 dB - 107 dB = 13 dB
Taking the logarithm function;
where;
= initial intensity
I = 19.95
I ≅ 20 pig callers
Answer:
because energy will be lost due to friction, sound, and heat (arguably similar to friction) and ENERGY MUST STAY THE SAME so it is IMPOSSIBLE for the ball to bounce higher than when dropped!
Answer:
(a) Most reactive 
Metal B
Metal D
Metal A
Least reactive Metal C
(b) (i) Bubbles should form very slowly
(ii) No reaction takes place
Explanation:
(a) The given metals arranged in their order of reactivity are;
Most reactive Metal B
Metal D
Metal A
Least reactive Metal C
The other of reactivity is based on the nature of their reactivity of the metals in air
(b) (i) Based on the reactivity of the metals in air, whereby metal A reacts very slowly and an oxide is formed, we have that, based on the reactivity of the metal A, when mixed with dilute hydrochloric acid, bubbles should form very slowly
(ii) Similarly, given that metal C is unreactive, we have that when small pieces of metal C are added to dilute hydrochloric acid, no reaction takes place.
Answer:
n = 1.4
Explanation:
Given,
R1 = 18 cm, R2 = -18 cm
From lens makers formula
1/f = (n - 1)(1/18 + 1/18) = (n-1)/9
f = 9/(n-1)
Power, P = 1/f ( in m) = (n-1)/0.09
Now, this lens is in with conjunction with a concave mirror which then can be thought of as to be in conjunction with another thin lens
Power of concave mirror = P' = 1/f ( in m) = 2/R = 2/0.18 = 1/0.09
Net power of the combination = 2P + P' = 2(n-1)/0.09 + 1/0.09 = 1/0.05
n = 1.4
