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loris [4]
3 years ago
7

As an IT technician for your company, you have been notified that the Windows domain does not seem to be functioning properly. B

eing familiar with domains, you are fairly confident you know what the issue is. But just to be safe, you take the applicable time to gather additional information and to identify what, if anything, has changed.
Which of the following is the BEST next step?

A. Determine the appropriate fix.
B. Create a hypothesis.
C. Implement the fix.
D. Identify what has changed.
E. Gather information.
Physics
1 answer:
Tasya [4]3 years ago
3 0

Answer: Create a hypothesis

Explanation:

From the information given, information has been gathered and the identification to ascertain if there's a change. Then, an hypothesis has to be created in order to know what the problem is.

One has to carry out some research in order to know what went wrong and should also validate the hypothesis by consulting with ones peers. By doing this, the most likely causes of the issues will be gotten.

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A parallel-plate capacitor is made from two aluminum-foil sheets, each 5.9 cm wide and 5.6 m long. Between the sheets is a Teflo
Zarrin [17]

Answer:

1.98 x 10⁻⁷ F

Explanation:

w = width of the sheet = 5.9 cm = 0.059 m

L = length of the sheet = 5.6 m

Area of the sheet is given as

A = L w = (5.6) (0.059) = 0.3304 m²

d = distance between the sheets = 3.1 x 10⁻⁵ m

k = dielectric constant of teflon = 2.1

Capacitance is given as

C = \frac{k\epsilon _{o}A}{d}

C = \frac{(2.1)(8.85\times 10^{-12})(0.3304)}{3.1\times 10^{-5}}

C = 1.98 x 10⁻⁷ F

5 0
3 years ago
A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f
VashaNatasha [74]

Answer:

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34  

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 =  X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + (\frac{L-y}{2}) ) g sinФ - F2 x (\frac{L}{2}) x gsinФ = 0

plug in the  equations of F1 and F2 into the above equation and after simplification, we get:

(L^{2} - y^{2} ) . w = L^{2} . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

(L^{2} - y^{2} ) . w = L^{2} . 5/9 x w

Hence,

(L^{2} - y^{2} ) = \frac{5L^{2} }{9}

\frac{L^{2} - y^{2}  }{L^{2} } = \frac{5}{9}

Taking L^{2} common and solving for \frac{y}{L}, we will get

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34

8 0
3 years ago
RC time constant circuit if R 50 KOC-21 a TOSS c. 1.05 s . what is the expected RC value b. 10.55 d. 0.105 s
Afina-wow [57]

Answer:

Time constant of RC circuit is 0.105 seconds.

Explanation:

It is given that,

Resistance, R=50\ K\Omega=5\times 10^4\ \Omega

Capacitance, C=2.1\ \mu F=2.1\times 10^{-6}\ F

We need to find the expected time constant for this RC circuit. It can be calculated as :

\tau=R\times C

\tau=5\times 10^4\times 2.1\times 10^{-6}

\tau=0.105\ s

So, the time constant for this RC circuit is 0.105 seconds. Hence, this is the required solution.

7 0
3 years ago
What are dimensionless quantities??​
bearhunter [10]

Answer:

Characteristic numbers are dimensionless numbers used in fluid dynamics to describe a character of the flow. To compare a real situation with a small-scale model it is necessary to keep the important characteristic numbers the same. Names of these numbers were standardized in ISO 31, part 12.

Explanation:

7 0
3 years ago
Ayudaaaaa por favorrrrrrrr!
abruzzese [7]
Same for #1 I have to say the same thing
6 0
3 years ago
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