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kiruha [24]
2 years ago
10

The amino and carboxyl functional groups tend to form bases and acids by attracting or droppingA) an electron.B) a neutron.C) a

proton and an electron.D) a neutron and a proton
Physics
1 answer:
statuscvo [17]2 years ago
7 0

Explanation:

When a substance or chemical specie accepts a proton then an acid is formed. Whereas is a substance or chemical specie tends to lose a proton then a base is formed.

Chemical formula of amino group is -NH_{2} and it is able to lose a proton to form R-NH-R'. When it tends to gain a proton the it form R-NH_{3} compound.

Chemical formula of carboxylic group is -COOH. When it loses a proton it forms -COO^{-} ion and whenit gains an electron then it forms -O-C-OH_{2}.

Therefore, we can conclude that the amino and carboxyl functional groups tend to form bases and acids by attracting or dropping a proton.

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A glider is gliding through the air at a height of 416 meters with a speed of 45.2 m/s. The glider dives to a height of 278 mete
Verdich [7]

Answer:

<em>The glider's new speed is 68.90 m/s</em>

Explanation:

<u>Principle Of Conservation Of Mechanical Energy</u>

The mechanical energy of a system is the sum of its kinetic and potential energy. When the only potential energy considered in the system is related to the height of an object, then it's called the gravitational potential energy. The kinetic energy of an object of mass m and speed v is

\displaystyle K=\frac{1}{2}mv^2

The gravitational potential energy when it's at a height h from the zero reference is

U=mgh

The total mechanical energy is

M=K+U

\displaystyle M=\frac{1}{2}mv^2+mgh

The principle of conservation of mechanical energy states the total energy is constant while no external force is applied to the system. One example of a non-conservative system happens when friction is considered since part of the energy is lost in its thermal manifestation.

The initial conditions of the problem state that our glider is glides at 416 meters with a speed of 45.2 m/s. The initial mechanical energy is

\displaystyle M_1=\frac{1}{2}m(45.2)v_o^2+m(9.8)(416)

Operating in terms of m

\displaystyle M_1=1021.52m+4076.8m

\displaystyle M_1=5098.32m

Then we know the glider dives to 278 meters and we need to know their final speed, let's call it v_f. The final mechanical energy is

\displaystyle M_2=\frac{1}{2}mv_f^2+m(9.8)(278)

Operating and factoring

\displaystyle M_2=m(\frac{1}{2}v_f^2+2724.4)

Both mechanical energies must be the same, so

\displaystyle m(\frac{1}{2}v_f^2+2724.4)=5098.32m

Simplifying by m and rearranging

\displaystyle \frac{v_f^2}{2}=5098.32-2724.4

Computing

v_f=\sqrt{4747.84}=68.90\ m/s

The glider's new speed is 68.90 m/s

8 0
2 years ago
Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 20.0 m above water wit
melamori03 [73]

Answer:

Explanation:

Given that,

Height of the bridge is 20m

Initial before he throws the rock

The height is hi = 20 m

Then, final height hitting the water

hf = 0 m

Initial speed the rock is throw

Vi = 15m/s

The final speed at which the rock hits the water

Vf = 24.8 m/s

Using conservation of energy given by the question hint

Ki + Ui = Kf + Uf

Where

Ki is initial kinetic energy

Ui is initial potential energy

Kf is final kinetic energy

Uf is final potential energy

Then,

Ki + Ui = Kf + Uf

Where

Ei = Ki + Ui

Where Ei is initial energy

Ei = ½mVi² + m•g•hi

Ei = ½m × 15² + m × 9.8 × 20

Ei = 112.5m + 196m

Ei = 308.5m J

Now,

Ef = Kf + Uf

Ef = ½mVf² + m•g•hf

Ef = ½m × 24.8² + m × 9.8 × 0

Ef = 307.52m + 0

Ef = 307.52m J

Since Ef ≈ Ei, then the rock thrown from the tip of a bridge is independent of the direction of throw

7 0
2 years ago
Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p
worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

  • U1/U2=6
  • U1/U2=6

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy  is mathematically given as

U=\frac{-k*qQ}{r}

Therefore

The Equations of U1 and U2 is

For U1

U1=\frac{-k*q_1Q}{r}

For U2

U2=\frac{-k*q_1Q}{3*2r}

Since

U is a function of q and  q2=q1/3

Therefore

U1/U2=6

For Question 2

For U1

U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}

Therefore

U1/U2=6

For more information on this visit

brainly.com/question/23379286?referrer=searchResults

7 0
2 years ago
A student throws a 5.0 newton ball straight up. What is the net force on the ball at its maximum height
Alex787 [66]
Iodine is the answer to your question buddy 
4 0
2 years ago
An object is traveling at a constant velocity of 15 m/s when it experiences a constant acceleration of 3.5 m/s2 for a time of 11
Luda [366]

Vf=Vi+at=15m/s+(3.5m/s^2)(11s)=53.5m/s

7 0
2 years ago
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