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Yanka [14]
3 years ago
9

Which statement supports the giant-impact hypothesis of the moon’s formation

Physics
2 answers:
V125BC [204]3 years ago
7 0

Answer:

the moon lacks a sizeable iron core

Explanation:

Katen [24]3 years ago
5 0

Answer:

The moon lacks a sizable iron core is the answer

Explanation:

<h2>Got it correct on BRAINLY <----≡</h2>
You might be interested in
A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = \frac{I}{nqA}

Where;

n = number of free electrons per cubic meter

q =  electron charge

A =  cross-sectional area of the wire

<em>First let's calculate the number of free electrons per cubic meter (n)</em>

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M)          -------------(ii)

<em>Substitute the values of Nₐ, ρ and M into equation (ii) as follows;</em>

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

<em>Now let's calculate the drift electron</em>

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

<em>Substitute these values into equation (i) as follows;</em>

v = \frac{I}{nqA}

v = \frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0
4 years ago
INT Raindrops acquire an electric charge as they fall. Suppose a 2.0-mm-diameter drop has a charge of 12 pC; these are both very
horsena [70]

Answer:

W = 2.3 10² F_{e}

Explanation:

The force of the weight is

         W = m g

         

let's use the concept of density

         ρ= m / v

the volume of a sphere is

         V = \frac{4}{3} π r³

         V = \frac{4}{3} π (1.0 10⁻³)³

         V = 4.1887 10⁻⁹ m³

the density of water ρ = 1000 kg / m³

          m = ρ V

          m = 1000 4.1887 10⁻⁹

          m = 4.1887 10⁻⁶ kg

therefore the out of gravity is

          W = 4.1887 10⁻⁶ 9.8

          W = 41.05 10⁻⁶ N

now let's look for the electric force

           F_e = q E

           F_e = 12 10⁻¹² 15000

           F_e = 1.8 10⁻⁷ N

         

the relationship between these two quantities is

          \frac{W}{F_e} = 41.05 10⁻⁶ / 1.8 10⁻⁷

           \frac{W}{F_e} = 2,281 10²

             

             W = 2.3 10² F_{e}

therefore the weight of the drop is much greater than the electric force

3 0
3 years ago
In the modern quantum–based atomic theory, what is the name given to a particular space around the nucleus in which an electron
Marina86 [1]
<span>The correct answer is: Atomic orbital

Explanation:
In the modern quantum-based atomic theory, an atomic orbital is a mathematical function that is used to explain the wave-like behavior of an electron in an atom. In other words, the space around the nucleus in which an electron moves is expressed by the atomic orbital. Hence, the correct answer is Atomic orbital.</span>
6 0
3 years ago
Read 2 more answers
A centripetal force of 210 N acts on a 1,600-kg satellite moving with a speed of 5,500 m/s in a circular orbit around a planet.
Dafna1 [17]

Answer:

<h2>230476.19km</h2>

Explanation:

Step one:

given

Force F= 210N

mass m= 1600kg

velocity v=5500m/s

Step two

Required is the radius r

the expression for the force is

F_c = \frac{mv^2}{r}

substitute

210=1600*5500^2/r

cross multiply we have

210r=48400000000

divide both side by 210

r=230476190.476m

r=230476.19km

4 0
3 years ago
Calculate the approximate volume of a 0.600 mol sample of gas at 15.0°c and a pressure of 1.10 atm.
Law Incorporation [45]

Answer:

0.013 m^3

Explanation:

The problem can be solved by using the ideal gas equation:

pV=nRT

where

p=1.10 atm = 1.11\cdot 10^5 Pa is the gas pressure

V is the gas volume

n = 0.600 is the number of moles of the gas

R=8.314 J/(mol K) is the gas constant

T=15.0^{\circ}+273.15=288.15 K is the absolute temperature of the gas

Solving the equation for V, we find the volume of the gas:

V=\frac{nRT}{p}=\frac{(0.600 mol)(8.314 J/(mol K))(288.15 K)}{1.11\cdot 10^5 Pa}=0.013 m^3

6 0
4 years ago
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