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Nutka1998 [239]

3 years ago

10

Three wires are made of copper having circular cross sections. Wire 1 has a length l and radius r. Wire 2 has a length l and rad

ius 2r. Wire 3 has a length 2l and radius 3r. Which wire has the smallest resistance?

1 answer:

Alex73 [517]3 years ago

4 0

Explanation:

Below is an attachment containing the solution.

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the radius of the earth is 6380 km and the height of the mount everest is 8848m . if the value of the acceleration due to gravit

Pavlova-9 [17]

Explanation:

g = GM / R²

9.77 m/s² = GM / (6.38×10⁶ m + 8848 m)²

GM = 3.99×10¹⁴ m³/s²

At the surface of the earth:

g = GM / R²

g = (3.99×10¹⁴ m³/s²) / (6.38×10⁶ m)²

g = 9.80 m/s²

At the top of Mount Everest, the weight of a 50 kg object is:

F = mg

F = (50 kg) (9.77 m/s²)

F = 489 N

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4 years ago

What type of cars are electric and what kind of battery do they have

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lithium-ion battery.

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4 years ago

The work energy principle states that the change in kinetic energy of an object is equal to the net work done on the object. If

sattari [20]

Answer:

$v=\sqrt{2gh}\ m/s$

Explanation:

From work energy theorem

Work done by all forces = Change in kinetic energy

Lets take

m= mass of object

h=height from the ground surface

initial velocity of object = 0 m/s

The final velocity of object is v

Work done by gravitational force = m g . h

The final kinetic energy = 1/2 m v²

So

Work done by all forces = Change in kinetic energy

m g h = 1/2 m v² - 0

v² = 2 g h

$v=\sqrt{2gh}\ m/s$

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4 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:

<h2> $F=ma$ (1) </h2>

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:

<h2> $P=mg$ (2) </h2>

<u>For block 1: </u>

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

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

<h2> $P_{1}=14.7N$ (4) </h2>

<u>For block 2: </u>

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

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

<h2> $P_{2}=23.52N$ (6) </h2>

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

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

<h2> $P_{1}+P_{2}=(m_{1}+m_{2})a$ (8) </h2>

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