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Semenov [28]
3 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]3 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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Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces
Mandarinka [93]

Answer: 0.81\times 10^{16} beta particles

Explanation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 14.0 g

Molar mass = 137 g/mol

\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of cesium contains atoms =  6.023\times 10^{23}

0.102 moles of cesium contains atoms =  \frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}

The relation of atoms with time for radioactivbe decay is:

N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}

Where N_t =atoms left undecayed

N_0 = initial atoms

t = time taken for decay = 3 minutes

{t_{\frac{1}{2}}} = half life = 30.0 years = 1.577\times 10^7 minutes

The fraction that decays  :  1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}

Amount of particles that decay is  = 0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}

Thus 0.81\times 10^{16} beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0
3 years ago
What is it called when a reading is converted into computer language? computerizing surveying digitizing converting
disa [49]
Digitizing is the correct answer
3 0
3 years ago
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3 years ago
Why doesn’t a machine that increases force break the law of conservation of energy?
USPshnik [31]

Answer:

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3 years ago
A child on a tricycle is moving at a speed of 1.40 m/s at the start of a 2.25 m high and 12.4 m long incline. The total mass is
goblinko [34]

Answer:

The work done by the child as the tricycle travels down the incline is 416.96 J

Explanation:

Given;

initial velocity of the child, v_i = 1.4 m/s

final velocity of the child, v_f = 6.5 m/s

initial height of the inclined plane, h = 2.25 m

length of the inclined plane, L = 12.4 m

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frictional force, f_k = 41 N

The work done by the child is calculated as;

\Delta E_{mech} = W - f_{k} \Delta L\\\\W = \Delta E_{mech}  + f_{k} \Delta L\\\\W = (K.E_f - K.E_i) + (P.E_f - P.E_i) + f_{k} \Delta L\\\\W = \frac{1}{2} m(v_f^2 - v_i^2) + mg(h_f - h_i) + f_{k} \Delta L\\\\W = \frac{1}{2} \times 48(6.5^2 - 1.4^2) + 48\times 9.8(0-2.25) + (41\times 12.4)\\\\W = 966.96  \ - \ 1058.4 \ + \ 508.4\\\\W = 416.96 \ J

Therefore, the work done by the child as the tricycle travels down the incline is 416.96 J

5 0
3 years ago
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