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Lorico [155]
3 years ago
9

Which elements will bond ionically with barium such that the formula would be written as BaX2?

Physics
2 answers:
kondor19780726 [428]3 years ago
6 0

Explanation:

Barium is a metal and has excess of electrons. Thus, in order to become stable barium will donate its extra electrons to an electron deficient atom. As a result, there will be formation of ionic bond.

Non-metals have deficiency of electrons and thus, they will readily accept electrons from barium in order to complete their octet.  Mostly non-metals have a charge of -1.

Therefore, they will combine with barium to form a compound with formula BaX_{2}.

For example, when barium combines with fluorine, the formula will be BaF_{2}.

Hence, we can conclude that barium can combine with fluorine, iodine, chlorine, and bromine such that the formula could be written as BaF_{2}.

Cerrena [4.2K]3 years ago
3 0
Chlorine, iodine, and fluorine <span> will bond ionically with barium such that the formula would be written as BaX2.</span>
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9. A sports car travels on a straight road at 22.0 km/h and increases its speed to 57.0 km/h in
Varvara68 [4.7K]

Answer:

4.4 M/s

Explanation:

6 0
3 years ago
An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 60 cm long and has a mass of 3.8 kg, with the center of
Serggg [28]

Answer:

(a) τ = 26.58 Nm

(b) τ = 18.79 Nm

Explanation:

(a)

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

d₁ = perpendicular distance between ball and shoulder = 60 cm = 0.6 m

τ₁ = (29.4 N)(0.6 m)

τ₁ = 17.64 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of 60 cm = (0.4)(60 cm) = 24 cm = 0.24 m

τ₂ = (37.24 N)(0.24 m)

τ₂ = 8.94 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 17.64 Nm + 8.94 Nm

<u>τ = 26.58 Nm</u>

<u></u>

(b)

Now, the arm is at 45° below horizontal line.

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

42.42 cm = 0.4242 m

τ₁ = (29.4 N)(0.4242 m)

τ₁ = 12.47 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of (60 cm)(Cos 45°) = (0.4)(42.42 cm) = 16.96 cm = 0.1696 m

τ₂ = (37.24 N)(0.1696 m)

τ₂ = 6.32 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 12.47 Nm + 6.32 Nm

<u>τ = 18.79 Nm</u>

3 0
2 years ago
In the simplified version of Kepler's third law, P 2 = a3, the units of the orbital period P and the semimajor axis a of the ell
Orlov [11]

Answer:

The units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.

Explanation:

P² = a³ is the simplified version of Kepler's third law which governs the orbital motion of large bodies that orbit around a star. The orbit of each planet is an ellipse with the star at the focal point.

Therefore, if you square the year of each planet and divide it by the distance that it is from the star, you will get the same number for all the other planets.

Thus, the units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.

8 0
3 years ago
If we connect a third bulb in our series circuit, say one with 4
Alexus [3.1K]

Answer:

The current will decrease.

Explanation:

When another bulb is added, the resistance is going to increase. Keep in mind that the current is inversely proportional to the resistance (<em>Ohm's law: R= </em><em>V</em><em>/</em><em>I</em><em> </em><em>).</em> Therefore when the resistance increase, the current running in the circuit will decrease.

6 0
2 years ago
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Semmy [17]
<span> In </span>space<span>, where there is no air, sound has no way to travel.</span>
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