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3 years ago

How do you put air knight in slingshock mode.

Engineering

1 answer:

Nesterboy [21]3 years ago

3 0

Answer:

I am not sure I am understanding plz more context

Explanation:

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A car has a steering wheel with a 15 inch diameter that takes 18 lbs of Effort force to move is

elena-14-01-66 [18.8K]

Answer: A first class lever in static equilibrium has a 50lb resistance force and 15lb effort force. The lever's effort force is located 4 ft from the fulcrum.

Explanation:

4 0

2 years ago

2. When it comes to selling their crop, what are 3 options a farmer has when harvesting their grain?

tiny-mole [99]

Answer:

Sell his crop, use his crop as food, and sell his crop

Explanation:

6 0

3 years ago

In a tensile test on a steel specimen, true strain = 0.12 at a stress of 250 MPa. When true stress = 350 MPa, true strain = 0.26

scZoUnD [109]

Answer:

The strength coefficient is $625$ and the strain-hardening exponent is $0.435$

Explanation:

Given the true strain is 0.12 at 250 MPa stress.

Also, at 350 MPa the strain is 0.26.

We need to find $(K)$ and the $(n)$ .

$\sigma =K\epsilon^n$

We will plug the values in the formula.

$250=K\times (0.12)^n\\350=K\times (0.26)^n$

We will solve these equation.

$K=\frac{250}{(0.12)^n}$ plug this value in $350=K\times (0.26)^n$

$350=\frac{250}{(0.12)^n}\times (0.26)^n\\ \\\frac{350}{250}=\frac{(0.26)^n}{(0.12)^n}\\ \\1.4=(2.17)^n$

Taking a natural log both sides we get.

$ln(1.4)=ln(2.17)^n\\ln(1.4)=n\times ln(2.17)\\n=\frac{ln(1.4)}{ln(2.17)}\\ n=0.435$

Now, we will find value of $K$

$K=\frac{250}{(0.12)^n}$

$K=\frac{250}{(0.12)^{0.435}}\\ \\K=\frac{250}{0.40}\\\\K=625$

So, the strength coefficient is $625$ and the strain-hardening exponent is $0.435$ .

5 0

3 years ago

Challenge:

goldenfox [79]

Answer:

what?

Explanation:

8 0

3 years ago

1 pts

kherson [118]

Vibrations felt through the floor are due to unbalanced wheels. It would most likely be rear because if it’s the front the steering whee would also vibrate

3 0

3 years ago