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jeka57 [31]
3 years ago
9

Which Two moon phases are directly opposite each other?

Engineering
2 answers:
liraira [26]3 years ago
6 0

Answer:   Full moon and the New Moon

Explanation:

Masja [62]3 years ago
3 0

Answer:New moon and full moon

Explanation:

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If you deposit today 11,613 in an account earning 8% compound interest, for how long should you invest the money in order to ear
Elanso [62]
Y = a (b)^t/p

y is total money

a is original amount

b is growth / decay factor

t is time

p is the frequency of every growth or decay

15131.76 = 11613 x 1.08^x

15131.76 / 11613 = 1.08^x

1.303… = 1.08^x

log1.303…. = xlog1.08

x = 3.43902165741 years
3 0
2 years ago
What different tests did the team perform to come up with a workable design?
I am Lyosha [343]
They ran different shapes and materials through a wind tunnel to see which shape and material would decrease energy output so that it takes in equal COthan it puts out.
5 0
3 years ago
.The war of the currents in the 1880's involved Thomas Edison and Nikola Tesla on a reality TV show stranded on an island. Each
natali 33 [55]

Answer:

True

Explanation:

Nikola Tesla defeated Thomas Edison in the AC/DC battle of electric current.

7 0
3 years ago
A well insulated turbine operates at steady state. Steam enters the turbine at 4 MPa with a specific enthalpy of 3015.4 kJ/kg an
Anarel [89]

Answer:

power developed by the turbine = 6927.415 kW

Explanation:

given data

pressure = 4 MPa

specific enthalpy h1 = 3015.4 kJ/kg

velocity v1 = 10 m/s

pressure = 0.07 MPa

specific enthalpy h2 = 2431.7 kJ/kg

velocity v2 = 90 m/s

mass flow rate = 11.95 kg/s

solution

we apply here  thermodynamic equation that

energy equation that is

h1 + \frac{v1}{2}  + q = h2 + \frac{v2}{2}  + w

put here value with

turbine is insulated so q = 0

so here

3015.4 *1000 + \frac{10^2}{2}  =  2431.7 * 1000 + \frac{90^2}{2}  + w

solve we get

w = 579700 J/kg = 579.7 kJ/kg

and

W = mass flow rate × w

W = 11.95 × 579.7

W = 6927.415 kW

power developed by the turbine = 6927.415 kW

7 0
3 years ago
Determine the design moment strength for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compressio
SVETLANKA909090 [29]

This question is incomplete, the complete question is;

Determine the design moment strength (ϕMn) for a W21x73 steel beam with a simple span of 18 ft when lateral bracing for the compression flange is provided at the ends only (i.e., Lb = 18 ft). Report the result in kip-ft.

Use Fy=50 ksi and assume Cb=1.0 (if needed).

Answer: the design moment strength for the W21x73 steel beam is 566.25 f-ft

Explanation:

Given that;

section  W 21 x 73 steel beam;

now from the steel table table for this section;

Zx = Sx = 151 in³

also given that; fy = 50 ksi and Cb = 1.0

QMn = 0.9 × Fy × Zx

so we substitute

QMn = 0.9 × 50 × 151

QMn = 6795 k-inch

we know that;

12inch equals 1 foot

so

QMn = 6795 k-inch / 12

QMn = 566.25 f-ft

Therefore the design moment strength for the W21x73 steel beam is 566.25 f-ft

7 0
3 years ago
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