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STatiana [176]
3 years ago
10

Start by setting aside money in your savings account for your goals and for emergencies. Then, fill in the other categories with

your remaining money. Be sure to allocate all of your money, so you have $0 left.
Engineering
1 answer:
Oduvanchick [21]3 years ago
3 0

Answer:i do not know money i had at first

Explanation:

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A and B connect the gear box to the wheel assemblies of a tractor, and shaft C connects it to the engine. Shafts A and Blie in t
azamat

Answer:

The couples are not all on one axis or plane for that matter but if the A and B connector had to be specified it would go by the yz axis diagonal to the x axis with a magnitude of about 15. The direction of the axis would be pointed up to the second quadrant. Hope this was helpful

Explanation:

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4 years ago
Pre Test
OverLord2011 [107]

Answer:

a is greater torgue

please make brainlist

8 0
3 years ago
A piston-cylinder assembly contains 2 lb of air at a temperature of 540 °R and a pressure of 1 atm. The air compressed to a stat
dalvyx [7]

Answer:

123.9 Btu

Explanation:

The energy balance on the air is:

∆E = E2 − E1 = ∆KE + ∆PE + ∆U = Q + W

ignore  ∆KE and ∆PE,

W = ∆U − Q = m(u2 − u1) − Q;                               (u2 − u1 = 51.94 Btu/lb)

ideal gas properties is attached

W = (2 lb)(143.98 − 92.04) Btu/lb − (− 20 Btu) = 123.9 Btu

u2 − u1 ≈ cv(T2 − T1) = (0.173 Btu/lb°R)(840 − 540) °R = 51.9 Btu/lb

6 0
3 years ago
8. What is the density of an object with a mass of 290.5 g and volume of 83 cm 3?​
Leni [432]
Answer:

The density would be 218.5
6 0
3 years ago
Given: A graphite-moderated nuclear reactor. Heat is generated uniformly in uranium rods of 0.05m diameter at the rate of 7.5 x
sineoko [7]

Answer:

The maximum temperature at the center of the rod is found to be 517.24 °C

Explanation:

Assumptions:

1- Heat transfer is steady.

2- Heat transfer is in one dimension, due to axial symmetry.

3- Heat generation is uniform.

Now, we consider an inner imaginary cylinder of radius R inside the actual uranium rod of radius Ro. So, from steady state conditions, we know that, heat generated within the rod will be equal to the heat conducted at any point of the rod. So, from Fourier's Law, we write:

Heat Conduction Through Rod = Heat Generation

-kAdT/dr = qV

where,

k = thermal conductivity = 29.5 W/m.K

q = heat generation per unit volume = 7.5 x 10^7 W/m³

V = volume of rod = π r² l

A = area of rod = 2π r l

using these values, we get:

dT = - (q/2k)(r dr)

integrating from r = 0, where T(0) = To = Maximum center temperature, to r = Ro, where, T(Ro) = Ts = surface temperature = 120°C.

To -Ts = qr²/4k

To = Ts + qr²/4k

To = 120°C + (7.5 x 10^7 W/m³)(0.025 m)²/(4)(29.5 W/m.°C)

To = 120° C + 397.24° C

<u>To = 517.24° C</u>

5 0
4 years ago
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