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1 year ago

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The following molecule can be synthesized from an epoxide and alkyllithium reagent, followed by aqueous workup. However, there a

re two ways in which this molecule can be formed. In one approach, the epoxide has a higher molecular weight than the epoxide used in the other approach. Draw both versions in the boxes below. Draw the carbon-lithium bond as a covalent bond.Part 1:Draw the combination of epoxide and alkyllithium reagent that uses a lower molecular weight epoxide.Part 2: Draw the combination of epoxide and alkyllithium reagent that uses a higher molecular weight epoxide.

1 answer:

Assoli18 [71]1 year ago

8 0

X. X

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A certain slide projector has a 100 mm focal length lens. How far away is the screen, if a slide is placed 103 mm from the lens

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Answer:

3.43 m

Explanation:

f = 100 mm

u = - 103 mm

Let v be the distance between the screen and the lens of the projector.

Use lens equation

1 / f = 1 / v - 1 / u

1 / 100 = 1 / v + 1 / 103

1 / v = 1 / 100 - 1 / 103

1 / v = (103 - 100) / (100 x 103)

1 / v = 3 / 10300

v = 3433.33 mm = 3.43 m

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1 year ago

Magnus has reached the finals of a strength competition. In the first round, he has to pull a city bus as far as he can. One end

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Answer:

The workdone is $W_d =-4400J$

Explanation:

The free body diagram is shown on the first uploaded image

From the question we are given that

The force is on the force gauge $F = 2750 N$

The distance that Magnus pulled the bus $d = 1.60m$

Generally the workdone by the tension force on Magnus is

$Workdone = Force * displacement \ in \ the \ direction \ of \ force$

$W_d = F * (-d)$

This negative sign show that is tension force is in the opposite direction to Magnus movement (i.e the movement of the bus )

Substituting value we have

$Workdone = - 2750 * 1.60$

$=-4400 J$

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2 years ago

4. Susan observed that different kinds and amounts of fossils were present in a cliff behind her house. She wondered why changes

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5,10,15,20,25,30, that's how much it should have been

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2 years ago

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You toss a rock of mass m vertically upward. Air resistance can be neglected. The rock reaches a maximum height h above your han

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Answer with Explanation:

We are given that

Mass of rock=m

Maximum height=h

a.At maximum height, velocity,v=0

We know that

$v^2=u^2-2gh$

$0+2gh=u^2$

$u^2=2gh$

Height,h=h/4

Again, $v'^2=u^2-2g\times \frac{h}{4}$

$v'^2=2gh-\frac{gh}{2}=\frac{4gh-gh}{2}=\frac{3gh}{2}$

$v'=\sqrt{\frac{3gh}{2}}=\sqrt{\frac{3\times 9.8 h}{2}}=3.83\sqrt h$

Where $g=9.8 m/s^2$

b.When height,h=3h/4

$v'^2=u^2-2gh$

$v'^2=2gh-2g\times \frac{3h}{4}=2gh-\frac{3gh}{2}=\frac{4gh-3gh}{2}=\frac{gh}{2}$

$v'=\sqrt{\frac{9.8h}{2}}=2.2\sqrt h$

$v'=2.2\sqrt h$

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2 years ago

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olya-2409 [2.1K]

Explanation:

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Formula for change in fringe shift is as follows.

n = $\frac{2lv^{2}}{\lambda c^{2}}$

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According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

l = 11 m

$\lambda = 5.9 \times 10^{-7} m$

c = $3.0 \times 10^{8}$ m/s

Hence, putting the given values into the above formula as follows.

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

= $\sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}$

= $2.41363 \times 10^{9} m/s$

Thus, we can conclude that velocity deduced is $2.41363 \times 10^{9} m/s$ .

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1 year ago