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likoan [24]
3 years ago
8

PLEASE HURRY!!!!! 25 POINTS !!!! Sebastian wants to write the chemical formula for phosphorus tetraoxide. He identifies the chem

ical symbols as P and O. Which additional step will he take in his process? identify the elements involved as potassium and oxygen add a subscript of 4 after the O write the chemical symbols, in order, as OP use the prefix tetra before oxygen
Physics
2 answers:
aksik [14]3 years ago
8 0

Answer:

add a subscript of 4 after the O

Explanation:

The chemical formula of a compound is said to contain the elements in that compound with the respective number of atoms in each element. When writing chemical formula of a compound, the chemical symbols of each comprising element must be known. In this case involving Phosphorus tetraoxide, the symbols of the elements in this compound has been identified as P for Phosphorus and O for oxygen.

After identification of elemental symbols in the compound, Sebastian must now allocate the number of atoms that each element contains. According to the name of the compound, TETRA preceding oxygen means four (4), hence, four will be added as a subscript to O. Hence, at the end, the chemical formula will be PO4.

Please note that, the order of the chemical formula will go as the name implies Phosphorus tetraoxide, so phosphorus symbol (P) comes first, followed by oxygen symbol (O)

drek231 [11]3 years ago
7 0

Answer:

add a subscript of 4 after the O

Explanation:

I just took the quiz on Edgenuity.

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3 years ago
A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the c
puteri [66]

Correct question:

A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?

Answer:

the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

Explanation:

Given;

length of solenoid, L= 0.35 m

diameter of the solenoid, d = 0.04 m

current through the solenoid, I = 5.0 A

magnetic field in the center of the solenoid, 2.8 x 10⁻² T

The number of turns per meter for the solenoid is calculated as follows;

B =\mu_o  I(\frac{N}{L} )\\\\B =  \mu_o  I(n)\\\\n = \frac{B}{\mu_o I} \\\\n = \frac{2.8 \times 10^{-2}}{4 \pi \times 10^{-7} \times 5.0} \\\\n = 4.5 \times 10^3 \ turns/m

Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

3 0
3 years ago
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Answer:

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2 years 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

     So

           \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

4 0
3 years ago
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