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1 year ago

The frequency of new cases of a disorder within a given time period is referred to as:______

Physics

1 answer:

artcher [175]1 year ago

7 0

Incidence is referred to as the frequency of new cases of a disorder in a given period of time.

<h3>What is incidence?</h3>

The prevalence of an illness over a specific time period can be calculated using the incidence metric, which measures disease prevalence. In light of this, the incidence of a disease is the quantity of newly discovered cases. An illness' incidence rate is calculated by dividing the number of people at risk for the illness by the number of new cases of the illness. Out of 200 women who participated in the study (and who were free of breast cancer at the start of the study period), five would be diagnosed with breast cancer during the course of a year. In this case, the incidence of breast cancer in this cohort would be 0.025. (Or 2,500 per 100,000 study years among women)

To know more about incidence of a disease visit:

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There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel

taurus [48]

Answer:

The total electric potential at mid way due to 'q' is $\frac{q}{4\pi\epsilon_{o}d}$

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say) = d

The electric potential at a distance d due to 'Q' is:

$V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}$

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

$V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Similar is the case with plate B:

$V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

$V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}$

$V_{total} = \frac{q}{4\pi\epsilon_{o}d}$

Now,

The Electric field due to charge Q at a distance is given by:

$\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}$

Now, if the charge q is mid way between the field, then distance is $\frac{d}{2}$ .

Electric Field at plate A, $\vec{E_{A}}$ at midway due to charge q:

$\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Similarly, for plate B:

$\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

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3 years ago

What is the slope of the line if the rise of a line on a distance versus-time graph is 900 meters and the run is 3 minutes?

Alexandra [31]

600/3 = 200
the slope is 200m/min

OR

600/ (3/60) =
600 x 60/3 =
600 x 20 = 12,000 meters per hour

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3 years ago

Brainliest if correct Question 7 of 10

OleMash [197]

Answer:

I would think A

Explanation:

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2 years ago

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How does a wind turbine transform mechanical energy into electrical energy?

Alex777 [14]

Answer:

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3 years ago

In a thundercloud there may be an electric charge of +40 C near the top of the cloud and -40 C near the bottom of the cloud. The

Ratling [72]

Answer:

Electric force, $F=-3.59\times 10^6\ N$

Explanation:

Given that,

Electric charge 1, $q_1=+40\ C$

Electric charge 2, $q_2=-40\ C$

Distance, $d=2\ km=2\times 10^3\ m$

To find,

The electric force between these two sets of charges.

Solution,

There exists a force between two electric charges and this force is called electrostatic force. It is equal to the product of electric charged divided by square of distance between them.

$F=k\dfrac{q_1q_2}{d^2}$

k is the electrostatic constant

$F=8.99\times 10^9\times \dfrac{40\times (-40)}{(2\times 10^3)^2}$

$F=-3.59\times 10^6\ N$

So, the electric force between these two sets of charges is $-3.59\times 10^6\ N$ .

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3 years ago