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Semenov [28]
2 years ago
8

) It is necessary to maintain a constant fluid volume in the body. Therefore, the body adjusts the water outputs to be equal to

the inputs. So, if an animal consumes 1.0 kg of water in feed and 5.0 kg of drinking water, produces 0.5 kg of metabolic water in the body, and excretes a total of 4.5 kg of water in feces and urine, calculate the amount (kg) of evaporative water losses of this animal (show your work).
Physics
1 answer:
loris [4]2 years ago
8 0

Answer:

Evaporative Water Loss = 2 kg

Explanation:

According to the given condition, the water entering the body must be equal to the water leaving the body. Therefore,

Water Entering the Body = Water Leaving the Body

Feed Water + Drinking Water + Metabolic Water = Urine Water + Evaporative Water Loss

using the given values:

1 kg + 5 kg + 0.5 kg = 4.5 kg + Evaporative Water Loss

Evaporative Water Loss = 1 kg + 5 kg + 0.5 kg - 4.5 kg

<u>Evaporative Water Loss = 2 kg</u>

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6 0
3 years ago
astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating o
devlian [24]

Answer:

The right answer is:

(a) 63.83 kg

(b) 0.725 m/s

Explanation:

The given query seems to be incomplete. Below is the attachment of the full question is attached.

The given values are:

T = 3 sec

k = 280 N/m

(a)

The mass of the string will be:

⇒ T=2 \pi\sqrt{\frac{m}{k} }

or,

⇒ m=\frac{k T^2}{4 \pi^2}

On substituting the values, we get

⇒     =\frac{280\times (3)^2}{4 \pi^2}

⇒     =\frac{280\times 9}{4\times (3.14)^2}

⇒     =68.83 \ kg

(b)

The speed of the string will be:

⇒  \frac{1}{2}k(0.4)^2=\frac{1}{2}k(0.2)^2+\frac{1}{2}mv^2

then,

⇒             v=\sqrt{\frac{k((0.4)^2-(0.2)^2)}{m} }

On substituting the values, we get

⇒                =\sqrt{\frac{280\times ((0.4)^2-(0.2)^2)}{63.83} }

⇒                =\sqrt{\frac{280(0.16-0.04)}{63.83} }

⇒                =\sqrt{\frac{280\times 0.12}{63.83} }

⇒                =0.725 \ m/s

4 0
3 years ago
PLEASE HELP WITH THIS QUESTION. ​
Sophie [7]

Answer:

#2 and #3 respectively

Explanation:

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3 years ago
After being accelerated through a potential difference of 5.33 kv, a singly charged carbon ion (12c) moves in a circle of radius
LekaFEV [45]
I'm stuck on the same question, as well :(
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If the particle has mass m, how fast must it be moving away from the Sun's center of mass to escape the gravitational influence
IgorLugansk [536]

Answer:

Explanation:

M = 1.989 x 10^30 kg

R = 6.96 x 10^8 m

G = 6.67 x 10^-11 Nm²/kg²

Let the velocity is v.

v=\sqrt{\frac{2GM}{R}}

v=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 1.989\times 10^{30}}{6.96\times 10^{8}}}

v = 6.17 x 10^5 m/s

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3 years ago
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