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noname [10]
3 years ago
12

What is the relationship between the distance of the fall (height) and the formation of splines and satellite drops?

Engineering
1 answer:
kondor19780726 [428]3 years ago
7 0
Mammas mama,aka Kim’s,smamsmsmmsmsmsmsmsmsmamsmmsms,sms,sms,msms,sms,sms,sms,sis,s,s,s
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Which is one of the aspects in PR game marketing?
mr_godi [17]
C, I took the test already.
7 0
3 years ago
Read 2 more answers
For a Cu-Ni alloy containing 53 wt.% Ni and 47 wt.% Cu at 1300°C, calculate the wt.% of the alloy that is solid and wt.% of allo
Zina [86]

Answer:

Hello your question is incomplete attached below is the complete question

answer: wt.% of alloy that is solid = 61.5%

             wt.% of allot that is liquid = 38.5%

Explanation:

To determine the wt.% of the alloy that is solid

= \frac{R}{R +S } * 100

=  \frac{53-45}{58-45} * 100  = 61.5%

To determine the wt.% of the alloy that is liquid

= \frac{S}{S+R} * 100\\

= \frac{58-53}{58-45} *100 = 38.5%

attached below is a free hand sketch as well

7 0
3 years ago
At what distance from a point charge of 8.0 μC would the electrical potential be 4.8 × 104 V? (ke = 8.99 × 109 N⋅m2/C2)
kotegsom [21]

Answer:

1.498 m

Explanation:

Electric potential due to a point charge V = K × Q / r

4.8 × 10 ⁴ V = 8.99 × 10⁹ N.m²/C² × 8 × 10⁻⁶ C / r

r = 8.99 × 10⁹ N.m²/C² × 8 × 10⁻⁶ C / 4.8 × 10 ⁴ V = 1.498 m

5 0
3 years ago
All machines have three fundamental hazards: moving parts, point of operation, and?
OlgaM077 [116]

Answer:

All machines have three fundamental hazards: moving parts, point of operation, and the power transmission.

Explanation:

The unit that supplies power to the machine is a critical hazard due to high energy sources being potential fatal if proper protocols are not followed. This is why lockout tagout (LOTO) measures are put in place in order to protect people while they work on equipment.

3 0
2 years ago
Air, at a free-stream temperature of 27.0°C and a pressure of 1.00 atm, flows over the top surface of a flat plate in parallel f
Morgarella [4.7K]

Answer:

Explanation:

Given that:

V = 12.5m/s

L= 2.70m

b= 0.65m

T_{ \infty} = 27^0C= 273+27 = 300K

T_s= 127^0C = (127+273)= 400K

P = 1atm

Film temperature

T_f = \frac{T_s + T_{\infty}}{2} \\\\=\frac{400+300}{2} \\\\=350K

dynamic viscosity =

\mu =20.9096\times 10^{-6} m^2/sec

density = 0.9946kg/m³

Pr = 0.708564

K= 229.7984 * 10⁻³w/mk

Reynolds number,

Re = \frac{SUD}{\mu} =\frac{\ SUl}{\mu}

=\frac{0.9946 \times 12.5\times 2.7}{20.9096\times 10^-^6} \\\\Re=1605375.043

we have,

Nu=\frac{hL}{k} =0.037Re^{4/5}Pr^{1/3}\\\\\frac{h\times2.7}{29.79\times 10^-63} =0.037(1605375.043)^{4/5}(0.7085)^{1/3}\\\\h=33.53w/m^2k

we have,

heat transfer rate from top plate

\theta _1 =hA(T_s-T_{\infty})\\\\A=Lb\\\\=2.7*0.655\\\\ \theta_1=33.53*2.7*0.65(127/27)\\\\ \theta_1=5884.51w

7 0
3 years ago
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