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3 years ago

______ are an idication that your vehicle may be developing a cooling system problem.

Engineering

1 answer:

iris [78.8K]3 years ago

8 0

Answer:

The temperature gauge showing that the vehicle has been running warmer or has recently began to have issues from overheating is an idication that your vehicle may be developing a cooling system problem.

Explanation:

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julsineya [31]

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6 0

3 years ago

10% A steel beam W18x76 spans 32 feet and is subjected to a Moment of 334 kips-ft. Find the load w on the beam. Determine the de

Lilit [14]

Answer:

w = 10.437 kips

deflection at 1/4 span 20.83\E ft

at mid span = 1.23\E ft

shear stress 7.3629 psi

Explanation:

area of cross section = 18*76

length of span = 32 ft

moment = 334 kips-ft

we know that

moment = load *eccentricity

334 = w * 32

w = 10.437 kips

deflection at 1/4 span

$\delta = \frac{wa^2b^2}{3EI}$

$= \frac{10.4375*8^2 *24^2}{3E \frac{BD^3}{12}}$

$=\frac{10.437 *8^2*24^2}{3E \frac{18*16^3}{12}}$

= 20.83\E ft

at mid span

$\delta = \frac{wl^3}{48EI}$

$= \frac{10.43 *32^3}{48 *E*\frac{18*16^3}{12}}$

$\delta = 1.23\E ft$

shear stress

$\tau = \frac{w}{A} = \frac{10.43 7*10^3}{18*76} =7.3629 psi$

6 0

3 years ago

Vehicles begin to arrive at a parking lot at 8:10 am at a constant rate of 6 veh/min until 8:25 am. There is no arrival from 8:2

tresset_1 [31]

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2.175 175 175

3.em dilisues

4.bye classmate magbabalik pa ako

8 0

3 years ago

Check Your Understanding: True Stress and Stress A cylindrical specimen of a metal alloy 47.7 mm long and 9.72 mm in diameter is

VARVARA [1.3K]

Answer:

The answer is "583.042533 MPa".

Explanation:

Solve the following for the real state strain 1:

$\varepsilon_{T}=\In \frac{I_{il}}{I_{01}}$

Solve the following for the real stress and pressure for the stable. $\sigma_{r1}=K(\varepsilon_{r1})^{n}$

$K=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n}$

Solve the following for the true state stress and stress2.

$\sigma_{r2}=K(\varepsilon_{r2})^n$

$=\frac{\sigma_{r1}}{[\In \frac{I_{il}}{I_{01}}]^n} \times [\In \frac{I_{i2}}{I_{02}}]^n\\\\=\frac{399 \ MPa}{[In \frac{54.4}{47.7}]^{0.2}} \times [In \frac{57.8}{47.7}]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.14046122)]^{0.2}} \times [In (1.21174004)]^{0.2}\\\\ =\frac{399 \ MPa}{[ In (1.02663509)]} \times [In 1.03915873]\\\\=\frac{399 \ MPa}{0.0114161042} \times 0.0166818905\\\\= 399 \ MPa \times 1.46125948\\\\=583.042533\ \ MPa$

4 0

3 years ago

A common process for increasing the moisture content of air is to bubble it through a column of water. The air bubbles are assum

Tamiku [17]

Answer:

See the attached file for the answers.

Explanation:

Find attached of the explanation

4 0

3 years ago