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ozzi
3 years ago
10

A person feels hungry even though he doesnot do any work give reason​

Physics
1 answer:
iren2701 [21]3 years ago
5 0

Answer:

See the explanation below.

Explanation:

The human body as well as that of any mammal uses a function known as metabolism. Which is able to convert energy from food into useful energy for different activities. These activities can be physical or bodily. The activities of the body are those necessary to maintain the body temperature above 36 [°C] this same temperature keeps the blood at sufficient temperature to facilitate the functions of pumping blood through the veins and arteries of our body.

In such a way that when the body feels hungry, this is a mechanism that tells us that our body requires energy from food, to maintain normal body energy levels. For this reason athletes need a large caloric intake of food, to be able to meet the standards of the competition and others.

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The Earth is 149.6 million km from the Sun. What is this distance also known as?
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It's as far as I remember one Astronomical Unit
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3 years ago
An object with a density of 0.85 g/cc is dropped into each of the two beakers shown below. Beaker 1 has a density of 0.5 g/cc. B
Oliga [24]

Answer:

The object will sink in the liquid in beaker 1.

The object will float in the liquid in beaker 2

Explanation:

The density of an object relative to the density of a fluid determines if the object floats or sink in a fluid. The density of a material is the measure of the amount of mass of that material packed into a unit volume of that material.

For the beaker 1, the liquid in this beaker has a density of 0.5 g/cc, which is lesser than the density of the object (0.85 g/cc). This means that the object will add more mass than there should be to the volume of the space it displaces within the field. This results in the object sinking in the fluid.

For beaker 2, the liquid in this beaker has a density of 1 g/cc, which is more than the density of the object (0.85 g/cc). This means that the object will add less mass than there should be to the volume of the space it displaces within the field. This results in the object floating in the fluid.

8 0
4 years ago
Directions: Fill in the blank with the correct word. Choose from the list of possible answers in the word bank below:
Alexxx [7]

Answer:

1).atoms (3). mixture. (5). Element

2). particles (4). molecules (6). suspension

Explanation:

(7). Homogeneous (8). Heterogeneous

(9). compound (10). solutions

8 0
3 years ago
Insert
nordsb [41]

A) 320 count/min

B) 40 count/min

C) 80 count/min, 11400 years

Explanation:

A)

The activity of a radioactive sample is the number of decays per second in the sample.

The activity of a sample is therefore directly proportional to the number of nuclei in the sample:

A\propto N

where A is the activity and N the number of nuclei.

As a consequence, since the number of nuclei is proportional to the mass of the sample, the activity is also directly proportional to the mass of the sample:

A\propto m

where m is the mass of the sample.

In this problem:

- When the mass is m_1 = 1 g, the activity is A_1=16 count/min

- When the mass is m_2=20 g, the activity is A_2

So we can find A2 by using the rule of three:

\frac{A_1}{m_1}=\frac{A_2}{m_2}\\A_2=A_1 \frac{m_2}{m_1}=(16)\frac{20}{1}=320 count/min

B)

The equation describing the activity of a radioactive sample as a function of time is:

A(t)= A_0 e^{-\lambda t} (1)

where

A_0 is the initial activity at time t = 0

t is the time

\lambda is the decay constant, which gives the probability of decay

The decay constant can be found using the equation

\lambda = \frac{ln2}{t_{1/2}}

where t_{1/2} is the half-life, which is the amount of time it takes for the radioactive sample to halve its activity.

In this problem, carbon-14 has half-life of

t_{1/2}=5700 y

So its decay constant is

\lambda=\frac{ln2}{5700}=1.22\cdot 10^{-4} y^{-1}

We also know that the tree died

t = 17,100 years ago

and that the initial activity was

A_0 = 320 count/min (value calculated in part A, corresponding to a mass of 20 g)

So, substituting into eq(1), we find the new activity:

A(17,100) = (320)e^{-(1.22\cdot 10^{-4})(17,100)}=40 count/min

C)

We know that a sample of living wood has an activity of

A=16 count/min per 1 g of mass.

Here we have 5 g of mass, therefore the activity of the sample when it was living was:

A_0 = A\cdot 5 = (16)(5)=80 count/min

Moreover, here we have a sample of 5 g, with current activity of A=20 count/min: it means that its activity per gram of mass is

A'=\frac{20}{5}=4 count/min

We know that the activity halves after every half-life: Here the activity has became 1/4 of the original value, this means that 2 half-lives have passed, because:

- After 1 half-life, the activity drops from 16 count/min to 8 count/min

- After 2 half-lives, the activity dropd to 4 count/min

So the age of the wood is equal to 2 half-lives, which is:

t=2t_{1/2}=2(5700)=11,400 y

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Hydrogen makes stars and stars spend most of their lifetime making helium.

The heavier elements are made when an old-age, high mass star explodes as a nova or supernova and then dies.

4 0
4 years ago
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