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Anestetic [448]

4 years ago

10

Which of the following is used as part of a four-wheel drive system? A. Front drive axles B. Transfer Case C. Front Drive Shaft

D. All of the above

1 answer:

xxMikexx [17]4 years ago

4 0

Answer:

D: All of the above.

Have a good day!

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One of the key characteristics of ________ sessions is that no idea should be immediately accepted or rejected. prototype alpha

Alex_Xolod [135]

Answer:

brainstorming

Explanation:

8 0

3 years ago

How many grams of perchloric acid, HClO4, are contained in 37.6 g of 70.5 wt% aqueous perchloric acid? How many grams of water a

kap26 [50]

Answer :

The mass of perchloric acid is, 26.5 grams.

The mass of water in the same solution is, 11.1 grams

Explanation :

As we are given that 70.5 wt % aqueous perchloric acid that means 70.5 grams of perchloric acid present in 100 grams of solution.

Now we have to determine the mass of perchloric acid in 37.6 grams of aqueous perchloric acid.

As, 100 grams of aqueous perchloric acid (solution) contains 70.5 grams of perchloric acid.

So, 37.6 grams of aqueous perchloric acid (solution) contains $\frac{37.6}{100}\times 70.5=26.5$ grams of perchloric acid.

Thus, the mass of perchloric acid is, 26.5 grams.

Now we have to determine the mass of water are in the same solution.

Total mass of solution = 37.6 g

Mass of perchloric acid = 26.5 g

Mass of water = Total mass of solution - Mass of perchloric acid

Mass of water = 37.6 g - 26.5 g

Mass of water = 11.1 g

Thus, the mass of water in the same solution is, 11.1 grams

4 0

3 years ago

On diesel engines, data from ________ sensors are commonly used to adjust exhaust gas recirculation (EGR) rates.

e-lub [12.9K]

Answer:

Air mass sensors is the right answer i think

Explanation:

3 0

2 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

"Write a statement that outputs variable numItems. End with a newline. Program will be tested with different input values."

kirill [66]

Answer:

The solution code is written in Java.

System.out.println(numItems);

Explanation:

Java <em>println() </em>method can be used to display any string on the console terminal. We can use <em>println()</em> method to output the value held by variable <em>numItems.</em> The <em>numItems </em>is passed as the input parameter to <em>println()</em> and this will output the value of <em>numItems</em> to console terminal and at the same time the output with be ended with a newline automatically.

6 0

3 years ago