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Inessa [10]
3 years ago
12

What type of bond is formed when a metal from group 2 reacts with a nonmetal from group 6

Physics
1 answer:
-Dominant- [34]3 years ago
8 0

Answer:

The correct answer is Ionic Bond

Explanation:

Group 2 elements are majorly metals, which include Magnesium (12), calcium (20), and some other metals, while the nonmetal from group 6 includes Oxygen (8), sulfur (16). The nonmetal gain electrons from the metal.

When metals in group two combine with nonmetal in group 6 they form an Ionic bond which is a strong bond.

The remaining electrons in the outer shell are transferred  when the oppositely charged ions of both elements combined. Where the metal transfers electrons to the nonmetal. For example, Magnesium (Mg) + Oxygen (O) = MgO. Magnesium transfers its (2) outer electrons to oxygen.

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What happened to the speed of light if it travels from air into glass?
vovikov84 [41]

Answer:

it slows down

Explanation:

the answer

8 0
3 years ago
b. A force 100N makes an angle of Ø with the x axis and has a y component of 30 N. Find both the x and y component of the force
horsena [70]

Answer:

Explanation:

The y component of the force is 100 sinØ . But given that y component is 30N

so 100 sinØ = 30

sinØ = 0.3

Ø = 17.5°.

X component of force = 100 cosØ

= 100 cos17.5

= 95.35 N .

Y component of force = 30 N .

Angle Ø = 17.5°.

6 0
3 years ago
A football is kicked into the air from an initial height of 4 feet. The height, in feet, of the football above the ground is giv
kakasveta [241]

Answer: 0.5 seconds or 2.625 seconds

Explanation:

At t = 0, The ball is 4 ft above the ground.

The height of the football varies with time in the following way:

s(t) = -16 t² + 50 t + 4

we need to find the time in which the height would of the football would be 25 ft:

⇒25 = -16 t² + 50 t + 4

we need to solve the quadratic equation:

⇒ 16 t² - 50 t + 21 = 0

t = \frac{50 \pm \sqrt{50^2-4\times 16\times 21}}{2\times16}

⇒ t = 0.5 s or 2.625 s

Therefore, at t = 0.5 s or 2.625 s, the football would be 25 ft above the ground.

3 0
3 years ago
Read 2 more answers
Two particles oscillate in simple harmonic motion along a common straight-line segment of length 1.0 m. Each particle has a peri
igor_vitrenko [27]

Answer:

a) the particles are <em>0.217 m </em>apart

b) <em>the particles are moving in the same direction</em>.

Explanation:

a) The amplitude of the oscillations is A/2 and the period of each particle is

T = 1.5 s however, they differ by a phase of π/6 rad. Let the phase of the first particle be zero so that the phase of the second particle is π/6. So we can write the coordinates of each of the particles as,

x₁ = A/2 cos(ωt)

x₂ = A/2 cos(ωt + π/6)

we can write the angular frequency ω, as

ω = 2π / T

so,

x₁ = A/2 cos(2π / T)

x₂ = A/2 cos(2π / T + π/6)

Thus, the coordinates of the particles at t = 0.45 s are,

x₁ = A/2 cos((2π × 0.45) / 1.5)) = -0.155 A

x₂ = A/2 cos((2π × 0.45) / 1.5) + π/6) = -0.372 A

Their separation at that time is, therefore,

Δx = x₁ - x₂

    = -0.155 A + 0.372 A

    = 0.217 A

since A = 1 m

Thus,

<em>Δx  = 0.217 m</em>

<em></em>

<em></em>

b) In order to find their directions, we must take the derivatives at t = 0.45 s.

Therefore,

v₁ = dx₁ / dt

   = (-πA / T) sin(2πt / T)

   = -(π(1) / 1.5) sin(2π(0.45) / 1.5)

   = -1.99

and,

v₂ = dx₂ / dt

   = (-πA / T) sin((2πt / T) + π/6)

   = -(π(1) / 1.5) sin((2π(0.45) / 1.5) + π/6)

   = -1.40

Since both v₁ and v₂ are negative, this shows that <em>the particles are moving in the same direction</em>.

6 0
3 years ago
Use the drop-down menus to complete the statement.
VashaNatasha [74]

Answer: A is your answer i am sorry if i am wrong

Explanation:

he first PLCs were programmed with a technique that was based on relay logic wiring schematics. This eliminated the need to teach the electricians, technicians and engineers how to program a computer - but, this method has stuck and it is the most common technique for programming PLCs today.

7 0
3 years ago
Read 2 more answers
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