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3 years ago

When forming a bond, an atom that has 3 electrons in its second shell and a filled first shell will?

1 answer:

7 0

Macromolecule polymers are assembled by the connecting of monomers. An -OH group is detached from one monomer and a hydrogen atom is detached from an additional in a procedure named dehydration synthesis in the monomers bond. For every subunit supplementary to a macromolecule in which one water molecule is detached. Macromolecule polymers are broken down by breaking bonds among subunits. This procedure is named hydrolysis and is the opposite of dehydration. During hydrolysis the hydrogen atom is supplementary to one monomer and a hydroxyl cluster to the other and by breaking the covalent bond in the middle of the monomers.

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Alicia drew a pentagon with equal side length and equal angles.Then she added red lines of symmetry to her drawing. How many lin

IRISSAK [1]

The sum of angles in a regular pentagon is
180*(5 - 2) = 540°
Each internal angle is 540/5 = 108°.

Each vertex creates a line of symmetry to the midpoint of the opposite side,
as shown in the figure.

Answer: 5 lines of symmetry.

7 0

3 years ago

Which of the following is a form of potential energy?

ozzi

Answer: nuclear energy

Explanation: nuclear energy

8 0

3 years ago

The 6 kg block is then released and accelerates to the right, toward the 4 kg block. The surface is rough and the coefficient of

KatRina [158]

Answer:

v =3.41 m/s

Explanation:

given,

mass of block 1 = 6 Kg

mass of another block 2 = 4 Kg

coefficient of friction = 0.3

Assuming 6 Kg block is attached to the spring of spring constant 350 N/m

and distance between the two block is equal to 0.5 m

using formula

$U = \dfrac{1}{2}kx^2$

$U = \dfrac{1}{2}\times 350 \times 0.5^2$

U = 43.75 J

using conservation of energy

KE = U - f.d

where f is the frictional force acting

$\dfrac{1}{2}mv^2 = 43.75- \mu m g d$

$\dfrac{1}{2}\times 6 \times v^2 = 43.75- 0.3\times 6 \times 9.8 \times 0.5$

$v= \sqrt{11.643}$

v =3.41 m/s

4 0

4 years ago

A flock of ducks is trying to migrate south for the winter, but they keep being blown off course by a wind blowing from the west

Minchanka [31]

The ducks' flight path as observed by someone standing on the ground is the sum of the wind velocity and the ducks' velocity relative to the wind:

ducks (relative to wind) + wind (relative to Earth) = ducks (relative to Earth)

or equivalently,

$\vec v_{D/W}+\vec v_{W/E}=\vec v_{D/E}$

(see the attached graphic)

We have

ducks (relative to wind) = 7.0 m/s in some direction θ relative to the positive horizontal direction, or

$\vec v_{D/W}=\left(7.0\dfrac{\rm m}{\rm s}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)$

wind (relative to Earth) = 5.0 m/s due East, or

$\vec v_{W/E}=\left(5.0\dfrac{\rm m}{\rm s}\right)(\cos0^\circ\,\vec\imath+\sin0^\circ\,\vec\jmath)$

ducks (relative to earth) = some speed v due South, or

$\vec v_{D/E}=v(\cos270^\circ\,\vec\imath+\sin270^\circ\,\vec\jmath)$

Then by setting components equal, we have

$\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta+5.0\dfrac{\rm m}{\rm s}=0$

$\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v$

We only care about the direction for this question, which we get from the first equation:

$\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta=-5.0\dfrac{\rm m}{\rm s}$

$\cos\theta=-\dfrac57$

$\theta=\cos^{-1}\left(-\dfrac57\right)\text{ OR }\theta=360^\circ-\cos^{-1}\left(-\dfrac57\right)$

or approximately 136º or 224º.

Only one of these directions must be correct. Choosing between them is a matter of picking the one that satisfies both equations. We want

$\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v$

which means θ must be between 180º and 360º (since angles in this range have negative sine).

So the ducks must fly (relative to the air) in a direction 224º relative to the positive horizontal direction, or about 44º South of West.

8 0

3 years ago

Bonus Points. A ball was thrown straight up in the air from 1.2 meter above the grouind. After 3 seconds the ball returnes to th

nekit [7.7K]

Answer:

Explanation:

given that

Distance above the ground, s = 1.2 m

Time taken by the ball, t = 3 s

Velocity of the ball, v = 1.2/3 = 0.4 m/s

Maximum height reached by the ball is then given by the formula

H = v² / 2g

H = 0.4² / 2 * 9.8

H = 0.16 / 19.6

H = 0.0082 m or rather, 0.82 cm

5 0

3 years ago