All categories

2 years ago

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.

You might be interested in

A container of volume 0.6 m^3 contains 5.3 mol of argon gas at 24°C. Assuming argon behaves as an ideal gas, find the total inte

Vitek1552 [10]

Answer:

the internal energy of the gas is 433089.52 J

Explanation:

let n be the number of moles, R be the gas constant and T be the temperature in Kelvins.

the internal energy of an ideal gas is given by:

Ein = 3/2×n×R×T

= 3/2×(5.3)×(8.31451)×(24 + 273)

= 433089.52 J

Therefore, the internal energy of this gas is 433089.52 J.

5 0

2 years ago

A cart is driven by a large propeller or fan, which can accelerate or decelerate the cart. The cart starts out at the position x

mash [69]

Answer:

The acceleration of the cart is 1.0 m\s^2 in the negative direction.

Explanation:

Using the equation of motion:

Vf^2 = Vi^2 + 2*a*x

2*a*x = Vf^2 - Vi^2

a = (Vf^2 - Vi^2)/ 2*x

Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.

Let x = Xf -Xi

Where Xf is the final position of the cart and Xi the initial position of the cart.

x = 12.5 - 0

x = 12.5

The cart comes to a stop before changing direction

Vf = 0 m/s

a = (0^2 - 5^2)/ 2*12.5

a = - 1 m/s^2

The cart is decelerating

Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.

5 0

2 years ago

A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet

Gnom [1K]

Answer:

d. 37 °C

Explanation:

$m_{m}$ = mass of lump of metal = 250 g

$c_{m}$ = specific heat of lump of metal = 0.25 cal/g°C

$T_{mi}$ = Initial temperature of lump of metal = 70 °C

$m_{w}$ = mass of water = 75 g

$c_{w}$ = specific heat of water = 1 cal/g°C

$T_{wi}$ = Initial temperature of water = 20 °C

$m_{c}$ = mass of calorimeter = 500 g

$c_{c}$ = specific heat of calorimeter = 0.10 cal/g°C

$T_{ci}$ = Initial temperature of calorimeter = 20 °C

$T_{f}$ = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

$m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) + m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C$

6 0

2 years ago

(1) A typical person has a surface area of about 2m^2 and a layer of fat about 1cm thick (It can vary from 0.5 cm to 1.5 cm) . E

KATRIN_1 [288]

Answer:

1)H=714 Watts of energy is given out.

2)614.34 KCal energy is burnt in an hour in the room

3)In freezing conditions, 1337.09 KCal is spent in an hour.

Explanation:

The formula for thermal conductivity states that the rate of heat loss H is as follows:

H= $\frac{KA (T_{1}-T_{2})}{x}$ where,

K=Coefficient of thermal conductivity

A=Area of Cross Section

H=Rate of heat loss

x=Thickness of the material

$T_{1}-T_{2}$ =Temperature difference between the two surfaces of the body in context

$T_{1}$ =37°C ie. Body Temperature

$T_{2}$ =20°C ie. Room Temperature

For first case, $T_{1}-T_{2}$ =17°C or 17°F

K for fat= 0.21 $Wm^{-1}K^{-1}$

A=2 $m^{2}$

x=0.01 m

H= $\frac{0.21 * 2* 17}{0.01}$

H=714 Watts of energy is given out.

2) $Power$ × $Time(in seconds) = Energy$

Energy= 714 × 60 × 60=2570400 J

2570400 J=614.3403442 KCal

Therefore, 614.3403442 KCal energy is burnt in an hour.

3) Only the temperature difference becomes 37°C from 17°C , rest everything remains the same, therefore the energy will also vary due to the temperature factor and more energy will be spent as in freezing climate $T_{2}=0$ °C

E= $\frac{614.34*37}{17}$ =1337.09 KCal

In freezing conditions, 1337.09 KCal is spent in an hour.

8 0

2 years ago

Part A

irina [24]

Answer:

v' = -18 m/s

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

$p_{o} = p_{f} (1)$

The initial momentum can be expressed as follows (taking as positive the initial direction of the ball):

$m_{b} * v_{b} -M_{c}*V_{c} = m_{b} * 18 m/s + (-M_{c}* 20 m/s) (2)$

The final momentum can be expressed as follows (since we know that v'b is opposite to the initial vb):

$-(m_{b} * v'_{b}) + M_{c}*V'_{c} (3)$

If we assume that Mc >> mb, we can assume that the car doesn't change its speed at all as a result of the collision, so we can replace V'c by Vc in (3).
So, we can write again (3) as follows:

$-(m_{b} * v'_{b}) +(- M_{c}*V_{c}) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (4)$

Replacing (2) and (4) in (1), we get:

$m_{b} * 18 m/s + (-M_{c}* 20 m/s) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (5)$

Simplifying, and rearranging, we can solve for v'b, as follows:
$v'_{b} = -18 m/s (6)$ , which is reasonable, because everything happens as if the ball had hit a wall, and the ball simply had inverted its speed after the collision.

3 0

1 year ago