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2 years ago

Explain why protons attached to nitrogen (i.e. n-h show broad peaks in the 1h nmr and do not couple to other protons.

1 answer:

7 0

The element Nitrogen is usually present in almost all proteins, and it also forms strong bonds due to its ability to form a triple bond with its self, also to other elements.

<span>The nitrides use a variety of different oxidation and almost all the oxides forms gasses are at 25 degrees Celsius. The oxide of nitrogen are acidic, thus, easily attaching protons.</span>

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True or false ??

galben [10]

Well, gravity's acceleration is -9.81 m/s^2 so false.

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3 years ago

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How are sound energy and light energy similar?

ruslelena [56]

B. They both have a wavelength and a frequency

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3 years ago

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True or False? The superposition or overlapping of two waves always results in destructive interference between the different wa

san4es73 [151]

Answer:

Explanation:

False

When two waves overlap or superimpose over each other then they either undergo Constructive or destructive interference.

waves are the disturbance created by a force and add up to gives constructive interference when they are in the same line i.e. in the same phase.

When these disturbances are in the opposite phase then they superimpose to give destructive interference where the amplitude of the resulting wave will be much smaller as compared to original waves.

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3 years ago

The electric potential at a position located a distance of 20.7 mm from a positive point charge of 8.60×10-9C and 15.1 mm from a

max2010maxim [7]

Answer:

q2 = -4.35*10^-9C

Explanation:

In order to find the values of the second charge, you use the following formula:

$V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}$ (1)

V: electric potential = 1.14 kV = 1.14*10^3 kV

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1: charge 1 = 8.60*10^-9 C

q2: charge 2 = ?

r1: distance to the first charge = 20.7mm = 20.7*10^-3 m

r2: distance to the second charge = 15.1mm

You solve the equation (1) for q2, and replace the values of the other parameters:

$q_2=\frac{r_2}{k}[V-k\frac{q_1}{r_1}]=\frac{Vr_2}{k}-\frac{q_1r_2}{r_1}\\\\q_2=\frac{(1.14*10^3V)(15.1*10^{-3}m)}{8.98*10^9Nm^2/C^2}-\frac{(8.60*10^{-9}C)(15.1*10^{-3}m)}{20.7*10^{-3}m}\\\\q_2=-4.35*10^{-9}C$

The values of the second charge is -4.35*10^-9C

8 0

2 years ago

A playground slide is inclined 40°. If a boy with a mass of 32 kg slides down for -3meters. How much work is done by gravity on

telo118 [61]

Answer:

the work done by gravity on the boy is 604.62 J

Explanation:

Given;

distance the boy slides, d = 3 m

angle of inclination of the playground, θ = 40⁰

mass of the boy, m = 32 kg

The vertical height, h, above the ground through which the boy falls represents the height of the triangle which is the opposite side.

The distance through which the boy slides, d, represents the hypotenuse side of the right triangle.

$sin \theta = \frac{opposite }{hypotenuse} \\\\sin \theta = \frac{h}{d} \\\\h = dsin\theta\\\\h = 3 \times sin(40^0)\\\\h = 1.928 \ m$

The work done by gravity on the boy is calculated as;

W = P.E = mgh

= 32kg x 9.8m/s² x 1.928m

= 604.62 J

Therefore, the work done by gravity on the boy is 604.62 J

8 0

3 years ago