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Sedbober [7]
2 years ago
7

Changes that occur in the urinary system with aging include all of the following, EXCEPT

Physics
1 answer:
Sladkaya [172]2 years ago
6 0

Answer: c. increased sensitivity to ADH

Explanation:

a. a decline in the number of functional nephrons: With aging the loss of nephron occurs that can be detected by the age related decrease in the glomerular filteration rate.

b. a reduction in the GFR (glomerular filtration rate): The GFR tend to decline in older age even though there is no disease. These people are required to check with the GFR in future.

d. problems with the micturition reflex: With aging people experience problem of bladder control. This leads to leakage or incontinence of urine or urinary retention that is inability to empty the bladder.

e. loss of sphincter muscle tone: With age the sphincter tone may diminish. This results in loss of control and storage capacity. The rectal muscles or sphincter muscles get loose which lead to passage of stool before reaching the washroom.

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Answer:

60

Explanation:

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Consider the following spectrum where two colorful lines (A and B) are positioned on a dark background. The violet end of the sp
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Answer:

Explanation:

a )

This type of spectrum is called line emission spectrum . Because it consists of lines . It is emission spectrum because it is due to emission of radiation from a source .

b ) The wavelength of a photon  is inversely proportional to its energy .  Photon  due to transition between n = 1 and n = 3 will have higher energy than

that due to transition between n = 2 and n = 5 . So the later photon ( B)  will have greater wavelength or photon  due to transition between n = 2 and n = 5 will have greater wavelength .

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2 years ago
At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci
kozerog [31]

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

3 0
2 years ago
A "biconvex" lens is one in which both surfaces of the lens bulge outwards. Suppose you had a biconvex lens with radii of curvat
Stolb23 [73]

Answer: f=150cm in water and f=60cm in air.

Explanation: Focal length is a measurement of how strong light is converged or diverged by a system. To find the variable, it can be used the formula:

\frac{1}{f} = (nglass - ni)(\frac{1}{R1} - \frac{1}{R2}).

nglass is the index of refraction of the glass;

ni is the index of refraction of the medium you want, water in this case;

R1 is the curvature through which light enters the lens;

R2 is the curvature of the surface which it exits the lens;

Substituting and calculating for water (nwater = 1.3):

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f = \frac{30}{0.2} = 150

For air (nair = 1):

\frac{1}{f} = (1.5 - 1)(\frac{1}{10} - \frac{1}{15})

f = \frac{30}{0.5} = 60

In water, the focal length of the lens is f = 150cm.

In air, f = 60cm.

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