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3 years ago

Balance the equationP + 02 → P205

Physics

2 answers:

AlladinOne [14]3 years ago

6 0

Answer:

<u>4</u>P + <u>5</u>02 → <u>2</u>P205

Explanation:

Hence the above equation is balanced.

ivolga24 [154]3 years ago

3 0

Explanation:

$\underline{ \underline{ \large{ \text{Given \: chemical \: equation}}}} :$

$\tt{P \: + O _{2} \: \: \: \: ➞ \: \: P_{2} \: O_{5}}$

Here , Multiply P2O5 by 2 and O2 by 5 to equalize oxygen :

$\tt{P+ 5O_{2} \: \: ➞\: \: 2P_{2} \:O_{5}}$

Now , To equalize P atoms , Multiply P by 4

$\boxed{ \large{\tt{ 4P + 5O _{2} \: ➞ \: \: \: \: 2P_{2} \: O_{5}}}}$

Here , On the left side we have 4P and On the right side ,we have 4P. Again , On the left side , we have 10 O and On the right side , we have 10 O. And yippie , we got the balanced chemical equation !!!

Hope I helped ! ♡

Have a wonderful day / night ! ツ

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<h3>What is the magnitude of acceleration of the box?</h3>

The magnitude of the acceleration of the box is calculated by applying Newton's second law of motion as shown below;

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where;

m is the mass of the box
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1 year ago

A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp

Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

=> $F_x = Tcos \theta$

Substituting values

$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

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3 years ago

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3 years ago

Balance the equationP + 02 → P205 ​

Balance the equationP + 02 → P205