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Elden [556K]
3 years ago
7

Covalent bonds are characterized by...

Physics
2 answers:
tresset_1 [31]3 years ago
5 0

Answer:

a.) Sharing of electrons

Explanation:

A covalent bond is a form of chemical bonding which is characterized by the sharing of electrons between atoms.

Alik [6]3 years ago
4 0

Answer:

A. the sharing of electrons

Explanation:

Covalent bonds share electrons.

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Evelyn learns that a sound wave can be recorded electronically as an analog signal or as a digital signal. She investigates thes
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What is a mixture of rock and mineral fragments, organic matter, air, and water called?
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The answer should be soil
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A student launches a small 0.5 kg rocket with an initial speed of 30 m/s at an angle of 60°. Approximately how much time will th
Elena L [17]

Answer:

t=5.30s

Explanation:

The high reached by a proyectile in an uniformly accelerated motion is given by:

y=v_{0y}t-\frac{gt^2}{2}

The time that the rocket spends in the air is obtained for y = 0, since this is the time that the rocket travels before touching the ground. Recall that v_{0y}=v_0sin\theta. Solving for t:

0=(v_0sin\theta) t-\frac{gt^2}{2}\\\frac{gt}{2}=v_0sin\theta\\t=\frac{2v_0sin\theta}{g}\\t=\frac{2(30\frac{m}{s})sin(60^\circ)}{9.8\frac{m}{s^2}}\\t=5.30s

5 0
4 years ago
Let’s determine what lens is needed to correct the vision of a myopic eye. Assume that the far point of the myopic eye is 50 cmc
koban [17]

Answer:

Explanation:

Person suffering from myopia is unable to see objects beyond 50 cm  because far point of his vision is 50 cm . He requires a concave lens

( diverging lens ) to correct his vision .

After using the lens ,

ray of light coming from far off place will appear to be coming from 50 cm after refraction through lens so it will be focused on retina .

object distance u = ∝

image distance v =  - 50 cm

focal length f = ?

1 / v - 1 / u = 1 /f

- 1 / 50 - 1 / ∝ = 1 / f

- 1 / 50  = 1 / f

f = - 50 cm

focal length is negative so len is concave.

He requires  concave lens of focal length 50 cm .

7 0
3 years ago
If the range of the projectile is 4.3 m, the time-of-flight is T = 0.829 s, and air resistance is negligible, determine the foll
ankoles [38]

Answer

given,

range of the projectile = 4.3 m

time of flight = T = 0.829 s

v =\dfrac{d}{T}

v =\dfrac{4.3}{0.829}

     v = 5.19 m/s

vertical component of velocity of projectile

v_y = gt'

v_y = 9.8 \times {\dfrac{T}{2}}

v_y = 9.8 \times {\dfrac{0.829}{2}}

v_y =4.06\ m/s

a) Launch angle

 \theta = tan^{-1}(\dfrac{v_y}{v})

 \theta = tan^{-1}(\dfrac{4.06}{5.19})

    θ = 38°

b) initial speed of projectile

  v = \sqrt{v_x^2 + v_y^2}

  v = \sqrt{5.19^2 + 4.06^2}

         v = 6.59 m/s

c) maximum height reached by the projectile

     y_{max}=v_{avg}t'

     y_{max}=\dfrac{1}{2}v_y\dfrac{T}{2}

     y_{max}=\dfrac{1}{2}\times(g\dfrac{T}{2})\times\dfrac{T}{2}

     y_{max}=\dfrac{gT^2}{8}          

7 0
4 years ago
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