Answer
Explanation:
so the velocity is 39 feet per sec so the impact is 720 cm from the ground
take 720 * 39 sq
Answer:
Throughout the clarification section elsewhere here, the definition of the concern is mentioned.
Explanation:
- The stress-strain curve provides designers with a long list of critical parameters that are needed for utility development. Including capacity, longevity, elasticity, apparent viscosity, tension electricity, resilience, as well as flexural strength, a load-pressure assignment gave us several mechanical households at a certain point of operation. It also assists in manufacturing.
- During which the overarching force can inform us about the maximum energy either workload the substance will experience, which could also be inferred within the action of the stress-strain. The dynamic properties can be seen by pre maximum activity because it will be before even the maximum yield intensity as well as the posted maximum would display plastic behavior however after the peak becomes achieved, the natural frequencies continue to decline.
Answer:
You're answer would be "D. Perform all of the above"
Explanation:
Have a Blessed Day (Mark Brainliest!)
Answer:
the atomic radius of a lithium atom is 0.152 nm
Explanation:
Given data in question
structure = BCC
lattice constant (a) = 0.35092 nm
to find out
atomic radius of a lithium atom
solution
we know structure is BCC
for BCC radius formula is
/4 × a
here we have known a value so we put a in radius formula
radius =
/4 × a
radius =
/4 × 0.35092
radius = 0.152 nm
so the atomic radius of a lithium atom is 0.152 nm
Answer:
a. 81 kj/kg
b. 420.625K
c. 101.24kj/kg
Explanation:
![\frac{t2}{t1} =[\frac{p2}{p1} ]^{\frac{y-1}{y} }](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7Bt1%7D%20%3D%5B%5Cfrac%7Bp2%7D%7Bp1%7D%20%5D%5E%7B%5Cfrac%7By-1%7D%7By%7D%20%7D)
t1 = 360
p1 = 0.4mpa
p2 = 1.20
y = 1.13
substitute these values into the equation
![\frac{t2}{360} =[\frac{1.20}{0.4} ]^{\frac{1.13-1}{1.13} }](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7B360%7D%20%3D%5B%5Cfrac%7B1.20%7D%7B0.4%7D%20%5D%5E%7B%5Cfrac%7B1.13-1%7D%7B1.13%7D%20%7D)
![\frac{t2}{360} =[\frac{1.2}{0.4} ]^{0.1150}\\\frac{t2}{360} =1.1347](https://tex.z-dn.net/?f=%5Cfrac%7Bt2%7D%7B360%7D%20%3D%5B%5Cfrac%7B1.2%7D%7B0.4%7D%20%5D%5E%7B0.1150%7D%5C%5C%5Cfrac%7Bt2%7D%7B360%7D%20%3D1.1347)
when we cross multiply
t2 = 360 * 1.1347
= 408.5
a. the work required in the firs compressor
w=c(t2-t1)
c=1.67x10³
t1 = 360
t2 = 408.5
w = 1670(408.5-360)
= 1670*48.5
= 80995 J
= 81KJ/kg
b. 
n = 80%
t2 = 408.5
t1 = 360
0.80 = 408.5-360 ÷ t'2-360

cross multiply to get the value of t'2
0.80(t'2-360) = 48.5
0.80t'2 - 288 = 48.5
0.8t'2 = 48.5+288
0.8t'2 = 336.5
t'2 = 336.5/0.8
= 420.625
this is the temperature at the exit of the first compressor
c. cooling requirement
w = c(t2-t1)
= 1.67x10³(420.625-360)
= 1670*60.625
= 101243.75
= 101.24kj/kg