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3 years ago

Which of these is largest? A. asteroids B. comets C. meteoroids D. planets

Physics

2 answers:

lys-0071 [83]3 years ago

5 0

Planets are the largest among the list. Take Jupiter or Saturn for example.

Effectus [21]3 years ago

5 0

Answer: [D]: "planets" .
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Which of the following element has seven total valence electrons? Your answer: argon helium oxygen bromine

GalinKa [24]

Answer:

bromine

Explanation:

Any element in the halogen group will have seven valence electrons. These elements include fluorine, chlorine, bromine, iodine, and astatine

6 0

3 years ago

Small, slowly moving spherical particles experience a drag force given by Stokes' law: Fd = 6πηrv where r is the radius of the p

Dominik [7]

Answer:

Explanation:

At the time of a body achieving terminal velocity, the drag force becomes equal to the weight of the body less the buoyant force by the surrounding medium which can be represented by the following equation

$\frac{4\pi\times r^3(d-\rho)}{3} =6\pi\times n\times r\times v$

Where r is radius of the body , d is density of the material of the body σ is density of the medium and n is coefficient of viscosity of the medium and v is terminal velocity.

Simplifying

v = $\frac{2\times r^2(d-\rho)}{9\times n}$

Assuming the value of density of air as 1.225 kg/m³ and putting other given values in the formula we get

v = $[tex]\frac{2\times (1.2\times10^{-5})^2(2182-1.225)}{9\times 1.8\times10^{-5}}$ [/tex]

v = 387 x 10⁻⁵ m/s

Terminal velocity = 387 x 10⁻⁵ m/s

Time taken to fall a distance of 100 m

= $\frac{100}{387\times10^{-5}}$

= 2.6 x 10⁴ s.

5 0

3 years ago

Find the direction and magnitude of the net force exerted on the point charge q3 in the figure. Let q= +2.4 μC and d= 33cm.

kobusy [5.1K]

With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x $10^{11}$ N and 66 degrees

ELECTRIC FORCE (F)

F = $\frac{KQq}{d^{2} }$

Where K = 9 x $10^{9}$ N $m^{2}$ / $C^{2}$

The distance between $q_{1}$ and $q_{3}$ can be calculated by using Pythagoras theorem.

d = $\sqrt{33^{2} + 33^{2} }$

d = 46.7 cm = 0.467 m

For force $F_{1}$ , substitute all the parameters into the formula above

$F_{1}$ = (9 x $10^{9}$ x 3 x 1)/ $0.467^{2}$

$F_{1}$ = 2.7 x $10^{10}$ /0.218

$F_{1}$ = 1.24 x $10^{11}$ N

For force $F_{4}$ , substitute all the parameters into the formula above

$F_{4}$ = (9 x $10^{9}$ x 3 x 4)/ $0.33^{2}$

$F_{4}$ = 1.08 x $10^{11}$ /0.1089

$F_{4}$ = 9.92 x $10^{11}$ N

For force $F_{2}$ , substitute all the parameters into the formula above

$F_{2}$ = (9 x $10^{9}$ x 3 x 2)/ $0.33^{2}$

$F_{2}$ = 5.4 x $10^{10}$ /0.1089

$F_{2}$ = 4.96 x $10^{11}$ N

Summation of forces on Y component will be

$F_{y}$ = $F_{4}$ - $F_{1}$ Sin 45

$F_{y}$ = 9.92 x $10^{11}$ - 1.24 x $10^{11}$ Sin 45

$F_{y}$ = 9.04 x $10^{11}$ N

Summation of forces on X component will be

$F_{x}$ = $F_{2}$ - $F_{1}$ Cos 45

$F_{x}$ = 4.96 x $10^{11}$ - 1.24 x $10^{11}$ Sin 45

$F_{x}$ = 4.08 x $10^{11}$ N

Net Force = $\sqrt{F_{x} ^{2} + F_{y} ^{2} } }$

Net force = $\sqrt{(4.08*10^{11}) ^{2} + (9.04*10^{11}) ^{2} }$

Net force = 9.9 x $10^{11}$ N

The direction will be

Tan ∅ = $F_{y}$ / $F_{x}$

Tan ∅ = 9.04 x $10^{11}$ / 4.08 x $10^{11}$

Tan ∅ = 2.216

∅ = $Tan^{-1}$ (2.216)

∅ = 65.7 degrees

Therefore, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x $10^{11}$ N and 66 degrees approximately.

Learn more about electric Force here: brainly.com/question/4053816

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2 years ago

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When copper combines with oxygen to form copper oxide, the charge of the copper ion is

Archy [21]

The copper and oxygen and oxygen atoms are not ions, and the bonds are more covalent than they are ionic.

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3 years ago

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A girl throws a rock horizontally at 10 m/s from the top of a building, 22 m above street level. Assuming free fall conditions a

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