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djverab [1.8K]
4 years ago
8

Predict whether an element with a valence of 8 is mostly likely form ionic bonds ,covalent bonds or no chemical bonds with other

atoms explain

Physics
2 answers:
Fiesta28 [93]4 years ago
6 0
They are more likely not to form since they already have 8 valence electrons and don't require the need of other atoms to create a bond with
spin [16.1K]4 years ago
4 0
An element with a valence of 8 are most likely noble gases (He, Ne, Ar, etc) which means they're already stable so they dont need to undergo chemical bonds with another atom because they already have a full valence shell.
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The pressure exerted by 15m of liquid is 1500pa.The acceleration due to gravity g=10m/s^2.Calculate the density liquid.​
Natali [406]

Answer:

1500 divided by 150(15m x 10m/s^2) = 10

8 0
3 years ago
A river flows due east at 1.60 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v
Citrus2011 [14]

Answer:

part (a) v\ =\ 10.42\ at\ 81.17^o towards north east direction.

part (b) s = 46.60 m

Explanation:

Given,

  • velocity of the river due to east = v_r\ =\ 1.60\ m/s.
  • velocity of the boat due to the north = v_b\ =\ 10.3\ m/s.

part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are 90^o

Let 'v' be the velocity of the boat relative to the shore.

\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.

Let \theta be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore.\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o

part (b)

  • Width of the shore = w = 300m

total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river.\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m

7 0
3 years ago
We are able to walk forward because _____.
Diano4ka-milaya [45]
Cuz Of Newton’s third law, every actions there is an equal and opposite reaction.
7 0
3 years ago
Read 2 more answers
Consider a satelite in a low altitude orbit around the Earth. The gravitational acceleration felt by the satelite is very close
likoan [24]

Answer:

The orbital speed of the satellite around the earth in other to remain in perfect circular orbit in given as:

v = sqrt[(Ge*M)/R],

where Ge is the gravitational constant (Ge = 6.673 x 10^-11N/m2/kg2), M is the mass of the earth(m = 5.98 x 10^24kg), and R is the radius of the earth (R = 6.47 x 10^6m)

v = SQRT [ (6.673 x 10^-11 N m2/kg2) • (5.98 x 10^24 kg) / (6.47 x 10^6 m) ]

v = 7.85 x 10^3 m/s

Explanation:

For a satelite in a low altitude orbit around the Earth, the gravitational force is the only force acting of the said satellite keeping it is a circular orbit. To keep this satellite in perfect circular orbit, it must be moving in at a certain speed, which is dependent on the earth mass and radius. This speed can be evaluated from the expression of centripetal force(F = mv2/r). The centripetal force Fc on the satellite is equal to the gravitational force on the satellite from the earth(Fe). That is, (Ge*M*m)/R2 = (m*v2)/R, where M is mass of the earth, and m is the mass of the satellite. making v the subject of the formula, the equation become v = sqrt[(Ge*M)/R].

4 0
3 years ago
A 191 191 kg sculpture hangs from a horizontal rod that serves as a pivot about which the sculpture can oscillate. The sculpture
Studentka2010 [4]

Answer:

r = 0.31 m

Explanation:

Given that,

Mass of the sculpture, m = 191 kg

The sculpture's moment of inertia with respect to the pivot is, I=17.2\ kg-m^2

Frequency of oscillation, f = 0.925 Hz

Let r is the distance of the the pivot from the sculpture's center of mass. The frequency of oscillation is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{mgr}{I}}

r=\dfrac{4\pi^2f^2I}{mg}

r=\dfrac{4\pi^2\times 0.925^2\times 17.2}{191\times 9.8}

r = 0.31 m

So, the pivot is 0.31 meters from the sculpture's center of mass. Hence, this is the required solution.

4 0
3 years ago
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