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3 years ago

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1. Band saw lower wheel does not require a guard * true or false 2. Band saw upper guide should be adjusted to within 1/8" of th

e work piece * true or false 3. Find board & linear ft for 10 pieces of 4" x 4" x 8' *

1 answer:

Blizzard [7]3 years ago

4 0

Answer:

1. Band saw lower wheel does not require a guard * true or false 2. Band saw upper guide should be adjusted to within 1/8" of the work piece * true or false 3. Find board & linear ft for 10 pieces of 4" x 4" x 8' *

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A unidirectional E-Glass fiber-epoxy composite material contains 61% by volume E-Glass fibers stressed under isostrain condition

zalisa [80]

Answer:

The total load carried by the fiber will be "98%".

Explanation:

The given values are:

$V_{f}=0.61$

$V_{m}=1-V_{f}$

$=1-0.61$

$=0.39$

$E_{f}=10 \ Mpa$

$\sigma_{f}=0.35 \ Mpa$

$E_{m}=0.45 \ Mpa$ , $\sigma_{m}=9\times 10^{-3} \ Mpa$

As we know,

⇒ $E_{e}=fE_{f}+mE_{m}$

On putting the estimated values, we get

⇒ $=0.61\times 10+0.39\times 0.95$

⇒ $=6.27 \ Mpa$

Now,

⇒ $\sigma_{c}=f\sigma_{f}+m\sigma_{m}$

On putting the estimated values, we get

⇒ $=0.61\times 0.35+0.39\times 0.009$

⇒ $=0.217 \ Mpa$

Therefore,

The load carried by fiber,

$=\frac{f\sigma_{f}}{\sigma_{c}}$

$=\frac{0.35\times 0.61}{0.217}$

$=0.98$ i.e., 98%

4 0

4 years ago

When a tensile specimen is stretched in the plastic region to an engineering strain of 0.2, calculate the amount of cold work pe

Valentin [98]

Answer:

0.2 x 100

Explanation:

Engineering strain is the original crossection/original crossection

cold work percentage is

original crossection/original crossection x 100

4 0

2 years ago

Ronny has a hydraulic jack. The input force is 250 N, while the output force is 7,500 N. If the area of the pipe below the input

Ganezh [65]

Answer:

6 m²

Explanation:

application of fluid pressure according to Pascal's principle for the two pistons is given as:

$P_1=P_2$

Where P₁ is the pressure at the input and P₂ is the pressure at the output.

But P₁ = F₁ / A₁ and P₂ = F₂ / A₂

Where F₁ and F₂ are the forces applied at the input and output respectively and A₁ and A₂ are the area of the input pipe and output pipe respectively

Since, $P_1=P_2$

$\frac{F_1}{A_1} =\frac{F_2}{A_2}\\$

But A₁ = 0.2 m², F₁ = 250 N, F₂ = 7500 N. Substituting values to get:

$\frac{F_1}{A_1} =\frac{F_2}{A_2}\\\frac{250}{0.2}=\frac{7500}{A_2}\\ A_2=\frac{7500*0.2}{250} = 6m^2$

Therefore, the area of the pipe below the load is 6 m²

6 0

3 years ago

The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".

If we analyze the staritning point we see that the initial velocity can be founded like this:

$v_o =\omega r_{OIC}=\omega (0.15m)$

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

$H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af$

$0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)$

And if we integrate the left part and we simplify the right part we have

$1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega$

And if we solve for $\omega$ we got:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

8 0

3 years ago

How does the engine thermostat regulate the engine temperature?

Inga [223]

Answer:

It's job is to block the flow of coolant to the radiator until the engine has warmed up. Once the engine reaches its operating temperature (generally about 200 degrees F, 95 degrees C), the thermostat opens. By letting the engine warm up as quickly as possible, the thermostat reduces engine wear, deposits and emissions.

6 0

4 years ago