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Dovator [93]
3 years ago
8

The surface area of the earth's crust is 5.10*10^8km^2. The average thickness of the earth's crust is 35km. The mean density of

the earth's crust is 2.8g/cm^3. The most abundant element in the earth's crust is oxygen with an abundance of 4.55*10^3g/t. Calculate the total volume of the earth's crust.
Physics
1 answer:
Mars2501 [29]3 years ago
4 0

Answer:

The answer is V = 1.785 \times 10^{10}\ Km^3.

Explanation:

You are asking only for the total volume, so the important data here is the surface area of the earth's crust,

A = 5.10 \times 10^8\ Km^2

and the average thickness of the earth's crust,

h = 35\ Km.

Imagine that you can stretch the surface area as if it were a blanket, now if you want to calculate its volume, you just need to multiply its area by its thickness,

V = A*h = 1.785 \times 10^{10}\ Km^3.

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pshichka [43]

Answer:

The answer is B) Double slit

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Bird man is flying horizontally at a speed of 33m/s and a height of 86m. Bird man releases a turd directly above the start of th
Aliun [14]

Answer:

2838

Explanation:

8 0
3 years ago
Rank the six combinations of electric charges on the basis of the electric force acting on q1.
loris [4]

Answer:

Largest; options A & B

2nd largest; Option F

3rd Largest; Option D

Smallest; Options E & C

Explanation:

Looking at the charges, basically, the net force of the charge q1 is the sum of charges q2 and q3.

When 2 charges are opposite, the force will be attractive but if they are same, then the force will be repulsive.

Thus, the order of the forces is;

I) Largest: q1= +1nC, q2= -1nc, q3= -1nc and (q1= -1nC, q2= +1nc, q3= +1nc)

II) Second largest is: q1= +1nC, q2= -1nc, q3= +1nc

III) Third largest: q1= +1nC, q2= +1nc, q3= -1nc

IV) Smallest: (q1= -1nC, q2= -1nc, q3= -1nc) and (q1= +1nC, q2= +1nc, q3= +1nc)

6 0
3 years ago
You’re standing at the highest point on the Moon, 10,786 mm above the level of the Moon’s mean radius. You’ve got a golf club an
Anna71 [15]

Answer:

A)   v = 1,675 10³ m / s  , B)    r₂ = 11,673 10⁶ m

Explanation:

A) This exercise we must use Newton's second law, where the forces of gravity are the Moon

        F = m a

acceleration is centripetal

        a = v² / r

force is the force of universal attraction

         F = G m M / r²

we substitute

        G m M / r² = m v² / r

        v² = G M / r

distance

        r = R_moon + h

        r = 1.74 10⁶ +1.0786 10⁴

        r = 1,750786 10⁶ m

we calculate

        v = √ (6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶)

        v = √ (2,8052 10⁶)

        v = 1,675 10³ m / s

B) let's use energy conservation

    Starting point. In the mountain

          Em₀ = K + U = ½ m v² + G m M / r

    Final point. Where the speed is zero

          Em_{f} = U = G mM / r₂

           Em₀ = Em_{f}

           ½ m v² + G m M / r = G mM / r₂

           1 / r₂ = (½ v₂ + G M / r) / GM

let's calculate

 1 / r₂ = (½ (1,675 10³)² + 6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶) /(6.67 10⁻¹¹ 7.36 10²²)

           1 / r₂ = (1,4028 10⁶ + 2,805 10⁶) / 49.12 10¹¹

           1 / r₂ = 8.5664 10⁻⁷

            r₂ = 11,673 10⁶ m

6 0
3 years ago
What can pressure be measured by?
SOVA2 [1]
'Pressure' is  (force) / (area).

The only choice with those units is #1 .
5 0
4 years ago
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