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morpeh [17]
3 years ago
8

Use the concept that y = c, −[infinity] < x < [infinity], is a constant function if and only if y' = 0 to determine whethe

r the given differential equation possesses constant solutions. (Enter all constant solutions below. If there are no constant solutions, enter NONE.) 2xy' + 4y = 8

Engineering
1 answer:
attashe74 [19]3 years ago
5 0

Answer:

attached below

Explanation:

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A plumbed eyewash station is portable.
Hitman42 [59]
Plumbed stations are permanently connected to a source of potable water, whereas portable stations are self-contained gravity-fed units with their own flushing fluid that must be replaced after each use. ... Eyewash fluid must irrigate and flush both eyes simultaneously.
Hopefully this helped.
8 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
2 years ago
technician A says that in any circuit, electrical current takes the path of least resistance. technician B says that while this
ArbitrLikvidat [17]

Answer:

  technician A is correct

Explanation:

Technician B has circuit topologies confused. In a series circuit, there is only one path for electrical current to take. In a parallel circuit, the current will divide between paths in proportion to the inverse of their resistance. The least resistance path will have the most current.

Technician A is mostly correct.

3 0
3 years ago
What's the highest grade level that brainly accomodates
taurus [48]

Answer:

The highest grade level is college.

3 0
2 years ago
In using the drag coefficient care needs to be taken to use the correct area when determining the drag force. What is a typical
stealth61 [152]

Answer:

Explanation:

We know that Drag forceF_D

  F_D=\dfrac{1}{2}C_D\rho AV^2

Where

             C_D is the drag force constant.

                 A is the projected area.

                V is the velocity.

                ρ is the density of fluid.

Form the above expression of drag force we can say that drag force depends on the area .So We should need to take care of correct are before finding drag force on body.

Example:

 When we place our hand out of the window in a moving car ,we feel a force in the opposite direction and feel like some one trying to pull our hand .This pulling force is nothing but it is drag force.

6 0
3 years ago
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