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2 years 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]2 years ago

8 0

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CHEGG 42 mT magnetic field points due west. If a proton of kinetic energy 9 x 10-12 J enters this field in an upward direction,

alexdok [17]

Answer:

The magnitude of the Force is $F = 697 *10^{-15}N$ and the direction is South

Explanation:

From the question we are told that

The magnetic field point due west and since East point toward the positive x -axis $(i)$ then this magnetic field would be mathematically represented as

$\= B = 42(-i)mT = 42*10^{-3} (-i) T$

Now from the question we are told that the kinetic energy is

$KE = 9*10^{-12}J$

Now this kinetic energy can be mathematically represented as

$KE = \frac{1}{2}mv^2$

Where m is the mass of proton which has a general value of

$m = 1.67*10^{-27}kg$

Now making the subject of the formula

$v = \sqrt{\frac{KE}{0.5 * m} }$

Substituting values we have

$v = \sqrt{\frac{9*10^{-12}}{0.5 * 1.67*10^{-27}} }$

$= 10.37*10^7m/s$

Now from the question we are told that proton is moving upward which is in the positive z direction so the velocity of the proton would be in the positive

So the velocity would be

$\= v = 10.37*10^{7} \r k \ m/s$

Now the magnetic Force can be mathematically represented as

$\= F = q \= v * \= B$

Where q is the charge on the proton which has a general value of $q =1.6*10^{-19}C$

Now substituting the value

$\= F = 1.6*10^{-19 } * (10.37 *10^7) \r k * (42 *10^{-3})(-i)$

$= 697*10^{-15} J$

Now according to Fleming's left hand rule the direction of the magnetic force is south toward the negative Y - direction $(-j)$

So the force can be denoted as

$\= F = 697*10^{-15}(-j) N$

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2 years ago

the steel bed of a suspension bridge is 200m long at 20 C. If the extremes of temperature to which it might be exposed are -30 C

raketka [301]

Answer:

The steel bed will contract by 0.13 m, and expand by 0.052 m

Explanation:

For contraction,

α = ΔL/(LΔθ)..................... Equation 1

Where α = Linear expansivity of steel, ΔL = decrease in length/ Increase in length, L = Original length, Δθ = Change in temperature

make ΔL the subject of the equation

ΔL = α(LΔθ)................. Equation 2

Given: α = 13×10⁻⁶/C, L = 200 m, Δθ = -30-20 = -50 °C

Substitute into equation 2

ΔL = 13×10⁻⁶(200)(-50)

ΔL = -0.13 m

Similarly, For expansion,

Using equation 2

ΔL = α(LΔθ)

Given: α = 13×10⁻⁶/C, L = 200 m, Δθ = 40-20 = 20 °C

Substitute into equation 2

ΔL = 13×10⁻⁶(200)(20)

ΔL = 0.052 m.

Hence the steel bed will contract by 0.13 m, and expand by 0.052 m

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2 years ago

The gravitational field strength at a distance R from the center of moon is gR. The satellite is moved to a new circular orbit t

3241004551 [841]

Answer:

$g'=\frac{g__R}{4}$

Explanation:

Given:

gravitational field strength of moon at a distance R from its center, $g__R$

Distance of the satellite from the center of the moon, $h=2R$

<u>Now as we know that the value of gravity of any heavenly body is at height h is given as:</u>

$g'=g__{R}} \times \frac{R^2}{(2R)^2}$

$g'=\frac{g__R}{4}$

∴The gravitational field strength will become one-fourth of what it is at the surface of the moon.

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2 years ago

Not too long ago houses were protected from excessive currents by fuses rather than circuit breakers. sometimes a fuse blew out

olganol [36]

Answer: Resistance = $8.21 \times 10^{-8} \Omega$

The approximate diameter of a penny is, <em>d</em> = 20 mm

thickness of penny is, <em>L = </em> 1.5× $10^{-3}$ mm

The area of penny along circular face is, $A = \frac{\pi d^2}{4} =\frac{\pi (20 mm\frac{1 m}{1000 m})^2}{4}$

= 3.14× $10^{-4}$ m²

The resistivity of copper is <em>ρ</em> = 1.72 x 10-8 Ωm.

Resistance,

$R = \rho \frac{L}{A} = (1.72 \times 10^{-8} \Omega m)\frac{1.5 \times 10^{-3} m}{3.14 \times 10^{-4} m^2} = 8.21 \times 10^{-8} \Omega$

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2 years ago

Kara Less was applying her makeup when she drove into South's busy parking lot last Friday morning. Unaware that Lisa Ford was s

exis [7]

Answer

given,

Mass of Kara's car = 1300 Kg

moving with speed = 11 m/s

time taken to stop = 0.14 s

final velocity = 0 m/s

distance between Lisa ford and Kara's car = 30 m

a) change in momentum of Kara's car

Δ P = m Δ v

$\Delta P = m (v_f-v_i)$

$\Delta P = 1300 (0 - 11)$

Δ P = - 1.43 x 10⁴ kg.m/s

b) impulse is equal to change in momentum of the car

I = - 1.43 x 10⁴ kg.m/s

c) magnitude of force experienced by Kara

I = F x t

I is impulse acting on the car

t is time

- 1.43 x 10⁴= F x 0.14

F = -1.021 x 10⁵ N

negative sign represents the direction of force

8 0

2 years ago

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