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4 years ago

14

Determine for which class of lever the output force is always greater than the input force. for which class is the output force

always less than the input force?

2 answers:

umka2103 [35]4 years ago

5 0

The 3rd class lever is the <span>output force always less than the input force, becuase its mechanical advantage is always less than 1. this also due that in a 3rd class lever the effort arm is shorter than the load arm, that is why the output is lower than the input force. but 3rd class lever is a speed multiplier lever</span>

Ad libitum [116K]4 years ago

4 0

Answer:

<u>Second Class of lever:</u> The second class of the lever is considered the most efficient or have the most mechanical advantage in conveying more productivity or output while investing less input into the system.

Explanation:

<u>The second Class of lever:</u>

The second type of lever has the load,L in between the effort,E and the fulcrum,F. As the effort moves a larger distance to move the load a small distance. As there is more productivity or mechanical output in response to less input to the system.

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the musical note A above middle C has a frequency of 440 Hz. If the speed of the sound is known to be 350 m/s, what is the wavel

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3 years ago

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The distance from earth to a NASA satellite traveling through the solar system is 6.0 × 109m. A command is sent to the satellite

alexandr1967 [171]

Answer:

the signal it takes 40 s to get the answer

Explanation:

As the satellite is in space the speed of radiation (waves9 is the speed of light in a vacuum, which is constant 3. 108 m / s, so we can use the uniform motion ratios to find the time.

v = d / t

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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

A volleyball is hit upward with an initial velocity of 6.0 m/s. Calculate the displacement of the volleyball when its final velo

Nimfa-mama [501]

Answer:

1.78 m upward

Explanation:

We can find the displacement of the volleyball by using the SUVAT equation:

$v^2 - u^2 = 2ad$

where, assuming upward as positive direction:

u = 6.0 m/s is the initial velocity

v = 1.1 m/s is the final velocity

a = g = -9.8 m/s^2 is the acceleration of gravity

d is the displacement

Solving the equation for d, we find:

$d=\frac{v^2-u^2}{2a}=\frac{1.1^2-6.0^2}{2(-9.8)}=1.78 m$

And since it is positive, the displacement is upward.

5 0

3 years ago

PLEASE HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

frozen [14]

A. Since it does not say the ball is moving your answer is A.

4 0

3 years ago