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larisa86 [58]
3 years ago
12

On some engines after torquing cylinder head fasteners you must

Engineering
1 answer:
loris [4]3 years ago
5 0

you must check that the head gasket is seated correctly and is not dislodged or slightly moved in some way or youl result in oil or coolant in the cylinders and loss of compression is possible to or a vacum leak in 1 or more cylinders

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Please perform 100 searches on Twitter using keyword "Lebron". Please print out the the number of tweets returned that were in E
Tom [10]

Answer:

what even is this????

5 0
3 years ago
Given the circuit at the right in which the following values are used: R1 = 20 kΩ, R2 = 12 kΩ, C = 10 µ F, and ε = 25 V. You clo
agasfer [191]

Answer:

a.) I = 7.8 × 10^-4 A

b.) V(20) = 9.3 × 10^-43 V

Explanation:

Given that the

R1 = 20 kΩ,

R2 = 12 kΩ,

C = 10 µ F, and

ε = 25 V.

R1 and R2 are in series with each other.

Let us first find the equivalent resistance R

R = R1 + R2

R = 20 + 12 = 32 kΩ

At t = 0, V = 25v

From ohms law, V = IR

Make current I the subject of formula

I = V/R

I = 25/32 × 10^3

I = 7.8 × 10^-4 A

b.) The voltage across R1 after a long time can be achieved by using the formula

V(t) = Voe^- (t/RC)

V(t) = 25e^- t/20000 × 10×10^-6

V(t) = 25e^- t/0.2

After a very long time. Let assume t = 20s. Then

V(20) = 25e^- 20/0.2

V(20) = 25e^-100

V(20) = 25 × 3.72 × 10^-44

V(20) = 9.3 × 10^-43 V

8 0
3 years ago
1. (15) A truck scale is made of a platform and four compression force sensors, one at each corner of the platform. The sensor i
Elanso [62]

Answer:

a). 139498.24 kg

b). 281.85 ohm

c). 10.2 ohm

Explanation:

Given :

Diameter, d = 22 m

Linear strain, $\epsilon$ = 3%

                        = 0.03

Young's modulus, E = 30 GPa

Gauge factor, k = 6.9

Gauge resistance, R = 340 Ω

a). Maximum truck weight

σ = Eε

σ = $0.03 \times 30 \times 10^9$

$\frac{P}{A} =0.03 \times 30 \times 10^9$

$P = 0.03 \times 30 \times 10^9\times \frac{\pi}{4}\times (0.022)^2$

 = 342119.44 N

For the four sensors,

Maximum weight = 4 x P

                            =  4 x 342119.44

                            = 1368477.76 N

Therefore, weight in kg is $m=\frac{W}{g}=\frac{1368477.76}{9.81}$

                     m = 139498.24 kg

b). Change in resistance

k=\frac{\Delta R/R}{\Delta L/L}

$\Delta R = k. \epsilon R$    , since $\epsilon= \Delta L/ L$

$\Delta R = 6.9 \times 0.03 \times 340$

$\Delta R = 70.38 $ Ω

For 4 resistance of the sensors,

$\Delta R = 70.38 \times 4 = 281.52$ Ω

c). $k=\frac{\Delta R/R}{\epsilon}$

If linear strain,

$\frac{\Delta R}{R} \approx \frac{\Delta L}{L}$  , where k = 1

$\Delta R = \frac{\Delta L}{L} \times R$

$\Delta R = 0.03 \times 340$

$\Delta R = 10.2 $ Ω

4 0
3 years ago
A flow of 100 mgd is to be developed from a 190-mi^2 watershed. At the flow line the area's reservoir is estimated to cover 3900
yaroslaw [1]

Answer:

13-mi 27 acres

Explanation:

7 0
3 years ago
One unethical decision won't cause any harm.( true) or (false)
konstantin123 [22]

Answer:

true

Explanation:

7 0
3 years ago
Read 2 more answers
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