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2 years ago

7

The elements least likely to form bonds are found in what group

2 answers:

klasskru [66]2 years ago

8 0

It is in the noble gas group which has a full valence electron shell found in group 18

lubasha [3.4K]2 years ago

7 0

Answer:

Group 18

Explanation:

Group 18 comprises of Noble elements (Neon, Krypton, Argon etc). These elements have complete outer shell that is complete octet. According to octet rule, the elements which have less than 8 electrons in their valence shell tend to bond with another element in order to complete their outer shell configuration.

Thus, group 18 elements are least likely to form bonds.

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I need help, please!

marta [7]

NO. it is not balanced

7 0

2 years ago

a particle is moving along a circular path having a radius of 4 in such that its position as a function of time is given by thet

ANTONII [103]

Answer:

Explanation:

Given

radius of circular path $r=4\ in.$

Position is given by

$\theta =\cos 2t---1$

Differentiate 1 to angular velocity we get

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega =-2\sin 2t----2$

Differentiate 2 to get angular acceleration

$\frac{\mathrm{d} \omega }{\mathrm{d} t}=-2^2\cos 2t ---3$

Net acceleration is the vector summation of tangential and centripetal force

$a_t=\alpha \times r$

$a_t=-4\cos 2t\times 4=-16\cos 2t$

$a_r=\omega ^2\cdot r$

$a_r=(-2\sin 2t)^2\cdot 4$

$a_r=16\sin^2(2t)$

$a_{net}=\sqrt{a_r^2+a_t^2}$

$a_{net}=\sqrt{(16\sin ^2(2t)+(-16\cos 2t)^2}$

$a_{net}=\sqrt{256\cos ^2(2t)+256\sin ^4(2t)}$

6 0

2 years ago

When a voltage difference is applied to a piece of metal wire, a 5.0 mA current flows through it. If this metal wire is now repl

zheka24 [161]

Answer:

I = 21.13 mA ≈ 21 mA

Explanation:

If

I₁ = 5 mA

L₁ = L₂ = L

V₁ = V₂ = V

ρ₁ = 1.68*10⁻⁸ Ohm-m

ρ₂ = 1.59*10⁻⁸ Ohm-m

D₁ = D

D₂ = 2D

S₁ = 0.25*π*D²

S₂ = 0.25*π*(2*D)² = π*D²

If we apply the equation

R = ρ*L / S

where (using Ohm's Law):

R = V / I

we have

V / I = ρ*L / S

If V and L are the same

V / L = ρ*I / S

then

(V / L)₁ = (V / L)₂ ⇒ ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂

If

S₁ = 0.25*π*D² and

S₂ = 0.25*π*(2*D)² = π*D²

we have

ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)

⇒ I₂ = 4*ρ₁*I₁ / ρ₂

⇒ I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m

⇒ I₂ = 21.13 mA

5 0

3 years ago

The nucleus of an atom contains a. only neutrons b protons and neutrons

nata0808 [166]

B. protons and neutrons

3 0

2 years ago

Read 2 more answers

A 2.00-m rod of negligible mass connects two very small objects at its ends. The mass of one object is 1.00 kg and the mass of t

8090 [49]

Answer:

<h2> 4kg</h2>

Explanation:

Step one:

given

length of rod=2m

mass of object 1 m1=1kg

let the unknown mass be x

center of mass<em> c.m</em>= 1.6m

hence 1kg is 1.6m from the <em>c.m</em>

and x is 0.4m from the <em>c.m</em>

Taking moment about the <em>c.m</em>

<em>clockwise moment equals anticlockwise moments</em>

1*1.6=x*0.4

1.6=0.4x

divide both sides by 0.4 we have

x=1.6/0.4

x=4kg

The mass of the other object is 4kg

3 0

2 years ago