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allsm [11]
3 years ago
12

Consider an ideal spring, with spring constant k, which is oriented along an x-axis. one end of the spring is fixed, and the fre

e end is located at x = 0 when the spring is neither stretched nor compressed. an object having a mass m is attached to the free end of the spring. consider an action in which the free end of the spring is moved from an initial position xi to a final position xf. the work done by the spring (and on the object attached to the free end) during this action is given by
Physics
1 answer:
Blizzard [7]3 years ago
5 0
The spring is neither stretched nor compressed. an object having a mass m is attached to the free end of the spring. consider an action
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Light strikes a 5.0-cm thick sheet of glass at an angle of incidence in air of 50°. the sheet has parallel faces and the glass h
mestny [16]
<span>Answer: sin(incidence)/sin(refraction) = n_refraction/n_incidence sin(50) / sin(x) = 1.5 / 1 sin(50)/1.5 = sin(x) sin(x) = 0.511 x = 30.71o B] 50 degrees, same as the angle going in. You can show that by reversing the steps in A. sin(30.7)/sin(x) = 1/1.5 C] The glass is 5 cm thick. The reference angle = 30.7o Tan(30.7) = displacement / thickness Tan(30.7) = x / 5 5*sin(30.7) = x x = 2.97 cm which is the displacement.</span>
4 0
3 years ago
A 12.5843-gram sample of metal bromide, MBr4, was dissolved and, after reaction with silver nitrate, AgNO3, all of the bromide w
barxatty [35]

Answer:

zirconium

Explanation:

Given, Mass of AgBr(s) = 23.0052 g

Molar mass of AgBr(s) = 187.77 g/mol

The formula for the calculation of moles is shown below:

Moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles\ of\ AgBr= \frac{23.0052\ g}{187.77\ g/mol}

Moles\ of\ AgBr= 0.1225\ mol

The reaction taking place is:

MBr_4+4AgNO_3\rightarrow 4AgBr+M(NO_3)__4

From the reaction,

4 moles of AgBr is produced when 1 mole of MBr_4 undergoes reaction

1 mole of AgBr is produced when 1 / 4 mole of MBr_4 undergoes reaction

0.1225 mole of AgBr is produced when \frac {1}{4}\times 0.1225 mole of MBr_4 undergoes reaction

Moles of MBr_4 got reacted = 0.030625 moles

Mass of the sample taken = 12.5843 g

Let the molar mass of the metal = x g/mol

So, Molar mass of MBr_4 = x + 4 × 79.904 g/mol = 319.616 + x g/mol

Thus,

0.030625 = \frac{12.5843}{319.616 + x}

Solve for x,

we get, x = 91.2999 g/mol

<u>The metal shows +4 oxidation state and has mass of 91.2999 g/mol . The metals is zirconium.</u>

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3 years ago
What do psychologist investigate
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