HELPP ME IN PHYSICS +15 POINTS!! But a right answer not links or I’ll report. -

Electron grops could be considered which of the following

Maurinko [17]

Answer:

Electron groups could be considered as Lone pair electrons and bonded pairs of electrons.

Answer: Option D & B

Explanation:

The two or more electrons can be bonded by single bond, double bond, covalent bond of electrons can simply be lone pair of electrons. Unshared pair of electrons are generally termed as lone pair of electrons in an atom which are generally present in the outermost shell of atoms. Hence electron groups can be determined by bonded pairs and lone pairs of electrons.

I got this from another brainly user

5 0

2 years ago

A sensitive gravimeter at a mountain observatory finds that the free-fall acceleration is 0.0055 m/s2 less than that at sea leve

pshichka [43]

Gs*rs^2 = gm*rm^2
rm = rs*√gs/gm 
rm = 6370*√9.83/(9.83-0.009) = 6372.92 
mountain observatory is placed at an altitude worth 2920 m asl

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7 0

3 years ago

. If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect tim

inn [45]

Answer:

The appropriate response will be "Length must be increased by 0.012%".

Explanation:

The given values is:

ΔT = 5 s/day

Now,

⇒ $\frac{\Delta T}{T} =\frac{5}{24\times 60\times 60}$

On multiplying both sides by "100", we get

⇒ $\frac{\Delta T}{T}\times 100 =\frac{500}{24\times 60\times 60}$

⇒ $\frac{\Delta T}{T}\times 100=0.005787$ (%)

∵ $T=2\pi\sqrt{\frac{l}{g} }$

On substituting the values, we get

⇒ $\frac{\Delta T}{T}$ % = $\frac{1}{2}\times \frac{\Delta l}{l}$ %

On applying cross multiplication, we get

⇒ $\frac{\Delta l}{l}$ % = $2\times \frac{\Delta T}{T}$ %

⇒ = $2\times 0.05787$

⇒ = $0.011574$

⇒ = $0.012$ %

6 0

3 years ago

Determine the ratio of the resistivity of pure water to silver?

shutvik [7]

Silver is a very good conductor, this means its resistivity is very low (from table, we can check the precise value, which is $\rho_s = 1.6 \cdot 10^{-8} \Omega m$ ).

Pure water, instead, is a very bad conductor, this means its resistivity is very high, of order of $k \Omega \cdot m$ ( $10^3 \Omega m$ ). Even without knowing the precise value of the pure water resistivity, we can estimate the ratio between the pure water resistivity and the silver resistivity by comparing the two orders of magnitude:
$r= \frac{10^3 \Omega m}{10^{-8} \Omega m} \sim 10^{11}$

Therefore, we can say that the correct answer is
$3 \cdot 10^{11} : 1$

3 0

3 years ago

The 26-kg sphere c is released from rest when θ = 0∘ and the tension in the spring is f = 100 n

aleksandr82 [10.1K]

You are given the mass of a sphere that is 26 kg sphere and it is released from rest when θ = 0°. You are also given the force of the spring that is F = 100 N. You are asked to find the tension of the spring. Imagine that the sphere is connected to a spring. The spring exerts a tension and the spring exerts gravitational pull. This will follow the second law of newton.

T - F = ma
T = ma + F
T = 26kg (9.81m/s²) + 100 N
T = 355.06 N

5 0

3 years ago

HELPP ME IN PHYSICS +15 POINTS!! But a right answer not links or I’ll report. -_-