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3 years ago

How much energy is contained in the mass of a 60-kilogram person?

2 answers:

OverLord2011 [107]3 years ago

7 0

The energy in the body depends on the work that had done and the type of food he she had if it is in a 60 kg person / 40 kg person.

Ratling [72]3 years ago

5 0

I could tell you the energy if you can tell me the acceleration and height above earth's surface.

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Energy that is associated with the position or composition of an object is called

Pani-rosa [81]

That is potential energy.

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3 years ago

What are Newton’s 3 laws

Vladimir79 [104]

Newton's First Law of Motion:

I. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.

Newton's Second Law of Motion:

II. The relationship between an object's mass m, its acceleration a, and the applied force F is F = ma. Acceleration and force are vectors (as indicated by their symbols being displayed in slant bold font); in this law the direction of the force vector is the same as the direction of the acceleration vector.

Newton's Third Law of Motion:

III. For every action there is an equal and opposite reaction.

3 0

3 years ago

Read 2 more answers

Worth 30 points! PLEASE ANSWER ASAP!!!

Juli2301 [7.4K]

V: there would be no seasons and humanity would suffer

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3 years ago

Allen Aby sets up an Atwood Machine and wants to find the acceleration and the tension in the string. Please help him. Two block

Vitek1552 [10]

<u>Answers:</u>

In order to solve this problem we will use Newton’s second Law, which is mathematically expressed after some simplifications as:

This can be read as: The Net Force $F$ of an object is equal to its mass $m$ multiplied by its acceleration $a$ .

We will also need to <u>draw the Free Body Diagram of each block</u> in order to know the direction of the acceleration in this system and find the Tension $T$ of the string (<u>See figure attached). </u>

We already know<u> $m_{2}$ is greater than $m_{1}$ </u>, this means the weight of the block 2 $P_{2}$ is greater than the weight of the block 1 $P_{1}$ ; therefore <u>the acceleration of the system will be in the direction of $P_{2}$ </u>, as shown in the figure attached.

We also know by the information given in the problem that <u>the pulley does not have friction and has negligible mass</u>, and <u>the string is massless</u>.

This means that the tension will be the same along the string regardless of the difference of mass of the blocks.

Now that we have the conditions clear, let’s begin with the calculations:

1) Firstly, we have to find the weight of each block, in order to verify that block 2 is heavier than block 1.

This is done using equation (1), where the force of the weight $P$ is calculated using the <u>acceleration of gravity</u> $g=9.8\frac{m}{s^{2}}$ acting on the blocks:

<u>For block 1: </u>

$P_{1}=m_{1}g$ (3)

$P_{1}=1.5kg(9.8\frac{m}{s^{2}})$

<u>For block 2: </u>

$P_{2}=m_{2}g$ (5)

$P_{2}=2.4kg(9.8\frac{m}{s^{2}})$

Then, we are going to <u>find the acceleration $a$ of the whole system: </u>

$F_{r}=P_{1}+P_{2}$ (7)

Where the Resulting Force $F_{r}$ is equal to the sum of the weights $P_{1}$ and $P_{2}$ .

In the figure attached, note that $P_{1}$ is in opposite direction to the acceleration $a$ , this means it must <u>have a negative sing</u>; while $P_{2}$ is in the same direction of $a$ .

Here we only have to isolate $a$ from equation (8) and substitute the values according to the conditions of the system:

$-14.7N+23.52N=(1.5kg+2.4kg)a$

$8.82N=(3.9kg)a$

Then:

$a=\frac{8.82N }{3.9kg}$

<h2> $a=2.26\frac{m}{ s^{2}}$ </h2><h2>This is the acceleration of the system. </h2>

2) For the second part of the problem, we have to find the tension $T$ of the string.

We can choose either the Free Body Diagram of block A or block B to make the calculations, <u>the result will be the same</u>.

Let’s prove it:

For $m_{1}$

we see in the free body diagram that the <u>acceleration is in the same direction of the tension of the string</u>, so:

$F_{r}=T-P_{1}$ (9)

$T-P_{1}=m_{1}a$ (10)

$T-14.7N=(1.5kg)( 2.26\frac{m}{ s^{2}})$

Then;

<h2> $T=18.09N$ This is the tension of the string </h2><h2> </h2>

For $m_{2}$

we see in the free body diagram that the acceleration is in opposite direction of the tension of the string and must <u>have a negative sign,</u> so:

$F_{r}=T-P_{2}$ (9)

$T-P_{2}=m_{2}a$ (10)

$T-23.52N=(2.4kg)(-2.26\frac{m}{ s^{2}})$

Then;

<h2> $T=18.09N$ This is the same tension of the string </h2>

6 0

3 years ago

A flat sheet of paper of area 0.250 m2 is oriented so that the normal to the sheet is at an angle of 60 to a uniform electric fi

Harman [31]

Answer:

a. 1.75 Nm²/C

b. Yes.

Explanation:

a. Electric Flux is given as:

Φ = E*A*cosθ

Where E = electric flux

A = Surface area

Φ = 14 * 0.25 * cos60

Φ = 1.75 Nm²/C

b. Yes, the shape of the sheet will affect the Flux through it. This is because flux is dependent on area of the surface and the area is dependent on the shape of the surface.

4 0

4 years ago