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zysi [14]
3 years ago
15

Jillian can't bear to leave her home without washing all the linens and floors, vacuuming the carpets, and organizing everything

in her cupboards. Because she knows that her behavior is extreme, she has tried to go to work without doing all these things, but she is overcome with such overwhelming dread that she never makes it all the way to the office. Jillian probably has
a. obsessive-compulsive disorder (OCD).
b. generalized anxiety disorder.
c. a phobia.
d. panic disorder.
Physics
1 answer:
kirill115 [55]3 years ago
7 0
In order to make any conclusive diagnosis, I would need to bring Jillian
into my office, and conduct a long, slow, complete, private examination.

Where was I ?

Oh, yes.

As I started to say:  Your description of Jillian's behavior is consistent
with a diagnosis of OCD.
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Kyra and Pari are timing how long it takes for 1 g of sugar to dissolve in hot water. Kyra records a time of 24.3 seconds. Pari
Hoochie [10]
<h3>Answer:</h3><h3>we can say that:-</h3>
  1. A reading with more no of significant figures is considered to be more precise.
  2. Kyra recorded a reading of 24.3 sec. Since all non 0 digits are considered to be significant this reading has 3 significant figures.
  3. Pari recorded a reading of 24 sec. Since all non 0 digits are considered to be significant this reading has 3 significant figures.
<h3>hence we can say that kyra's reading has more significant figures nd so it is more precise.</h3>

7 0
2 years ago
A ball is projected horizontally from the top of a bertical building 25.0m above the ground level with an initial velocity of 8.
kirill115 [55]

Answer:

Solution given:

height [H]=25m

initial velocity [u]=8.25m/s

g=9.8m/s

now;

a. How long is the ball in flight before striking the ground?

Time of flight =?

Now

Time of flight=\sqrt{\frac{2H}{g}}

substituting value

  • =\sqrt{\frac{2*25}{9.8}}
  • =2.26seconds

<h3><u>the ball is in flight before striking the ground for 2.26seconds</u>.</h3>

b. How far from the building does the ball strike the ground?

<u>H</u><u>o</u><u>r</u><u>i</u><u>z</u><u>o</u><u>n</u><u>t</u><u>a</u><u>l</u><u> </u>range=?

we have

Horizontal range=u*\sqrt{\frac{2H}{g}}

  • =8.25*2.26
  • =18.63m

<h3><u>The ball strikes 18.63m far from building</u>. </h3>
7 0
2 years ago
DNA is coiled into chromosomes in a cell's
tia_tia [17]

Answer: Cell's nucleus

Explanation :

The full form of DNA is Deoxyribonucleic acid. The genetic information of a cell is organized in the DNA. It is inherited from parents by their children.

DNA is coiled into chromosomes in a cell's nucleus. It has a shape like a double helix. It is twisted in the form of spirals.

Hence, DNA is coiled into chromosomes in a cell's Nucleus.

6 0
3 years ago
Read 2 more answers
A confined aquifer with a porosity of 0.15 is 30 m thick. The potentiometric surface elevation at two observation wells 1000 m a
AlekseyPX

Answer:

Part (a) The flow rate per unit width of the aquifer is 1.0875 m³/day

Part (b) The specific discharge of the flow is 0.0363 m/day

Part (c) The average linear velocity of the flow is 0.242 m/day

Part (d) The time taken for a tracer to travel the distance between the observation wells is 4132.23 days = 99173.52 hours

Explanation:

Part (a) the flow rate per unit width of the aquifer

From Darcy's law;

q = -Kb\frac{dh}{dl}

where;

q is the flow rate

K is the permeability or conductivity of the aquifer = 25  m/day

b is the aquifer thickness

dh is the change in th vertical hight = 50.9m - 52.35m = -1.45 m

dl is the change in the horizontal hight = 1000 m

q = -(25*30)*(-1.45/1000)

q = 1.0875 m³/day

Part (b) the specific discharge of the flow

V = \frac{Q}{A} = \frac{q}{b} = -K\frac{dh}{dl}\\\\V = -(25 m/d).(\frac{-1.45 m}{1000 m}) = 0.0363 m/day

V = 0.0363 m/day

Part (c) the average linear velocity of the flow assuming steady unidirectional flow

Va = V/Φ

Φ is the porosity = 0.15

Va = 0.0363 / 0.15

Va = 0.242 m/day

Part (d) the time taken for a tracer to travel the distance between the observation wells

The distance between the two wells = 1000 m

average linear velocity = 0.242 m/day

Time = distance / speed

Time = (1000 m) / (0.242 m/day)

Time = 4132.23 days

        = 4132.23 days *\frac{24 .hrs}{1.day} = 99173.52, hours

4 0
2 years ago
The attraction will vary directly with the separation between the charges.
Burka [1]
No it won't. It'll vary inversely as the square of the separation.
4 0
2 years ago
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