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Oksana_A [137]
3 years ago
9

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

Physics
1 answer:
Leto [7]3 years ago
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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Whats the origin of all stars?<br> a) supernova<br> b) dwarfs<br> c) protostars<br> d) nebulae
Fudgin [204]
All stars originated from c., protostars. Protostars are the earliest phase of the process of star evolution.

-Starry Sky 

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5 0
3 years ago
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In a collision, a 15 kg object moving with a velocity of 3 m/s transfers some of its momentum to a 5 kg object. What would be th
Misha Larkins [42]

The key to solve this problem is the conservation of momentum. The momentum of an object is defined as the product between the mass and the velocity, and it's usually labelled with the letter p:

p=mv

The total momentum is the sum of the momentums. The initial situation is the following:

m_A=15,\quad v_A=3,\quad m_B=5,\quad v_B=0

(it's not written explicitly, but I assume that the 5-kg object is still at the beginning).

So, at the beginning, the total momentum is

p=m_Av_A+m_Bv_B=15\cdot 3+5\cdot 0=45

At the end, we have

m_A=15,\quad v_A=1,\quad m_B=5,\quad v_B=x

(the mass obviously don't change, the new velocity of the 15-kg object is 1, and the velocity of the 5-kg object is unkown)

After the impact, the total momentum is

p=m_Av_A+m_Bv_B=15\cdot 1+5\cdot x=15+5x

Since the momentum is preserved, the initial and final momentum must be the same. Set an equation between the initial and final momentum and solve it for x, and you'll have the final velocity of the 5-kg object.

4 0
4 years ago
40
insens350 [35]

Answer:

774.8 secs

Explanation:

distance(d)= speed(v)* time(t)

calculate speed:

refractive index = speed of light (c)/ speed of light in medium (v)

1.56 = 3*10^8*v

v=192307692.3 m/s

d = v *t

t = d/v

on substituting values:

t = 774.8 secs

4 0
3 years ago
What is the acceleration of a 10 kg mass pushed by a 5 N force?
yanalaym [24]
g-\ gravitational \ acceleration \\ g= \frac{G}{m} = \frac{5N}{10kg}=  \\ g=0,5 \frac{N}{kg}
8 0
4 years ago
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In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec
Bezzdna [24]

Answer with Explanation:

We are given that

r=0.053 nm=0.053\times 10^{-9} m

1 nm=10^{-9} m

Charge on proton,q=1.6\times 10^{-19} C

a.We have to find the electric  potential of the proton at the position of the electron.

We know that the electric potential

V=\frac{kq}{r}

Where k=9\times 10^9

V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

V=27.17 V

B.Potential energy of electron,U=\frac{kq_e q_p}{r}

Where

q_e=-1.6\times 10^{-19} c=Charge on electron

q_p=q=1.6\times 10^{-19} C=Charge on proton

Using the formula

U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

U=-4.35\times 10^{-18} J

8 0
3 years ago
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