Answer:
6.86 × 10²⁴ kg
Explanation:
The mass of the earth m = density of earth, ρ × volume of earth, V
m = ρV
The density of the earth, ρ = 5515 kg/m³ and since the earth is a sphere, its volume is the volume of a sphere V = 4πr³/3 where r = radius of the earth = 6.67 × 10⁶ m
Since m = ρV
m = ρ4πr³/3
So, substituting the values of the variables into the equation for the mass of the earth, m, we have
m = 5515 kg/m³ × 4π(6.67 × 10⁶ m)³/3
m = 5515 kg/m³ × 4π × 296.741 × 10¹⁸ m³/3
m = 5515 kg/m³ × 1189.9639π × 10¹⁸ m³/3
m = 6546105.64378π × 10¹⁸ kg/3
m = 20565197.400122 × 10¹⁸ kg/3
m = 6855065.8 × 10¹⁸ kg
m = 6.8550658 × 10²⁴ kg
m ≅ 6.86 × 10²⁴ kg
a.the amount of sunlight increases.
Explanation:
As a submarine rises to the surface, the change it encounters that is true from the given options is that the amount of sunlight increases.
The bottom of the ocean is dark and receives little to no sunlight due to the scattering of the rays by ocean water.
- As the submarine rises, the volume of water column on it decreases and the pressure on it decreases too.
- Also, the temperature rises steadily to the surface.
learn more:
Heat and temperature brainly.com/question/914750
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The magnitude (in N) of the force she must exert on the wrench is 150.1 N.
<h3>
Force exerted by the wrench</h3>
The force exerted by the wrench is calculated using torque formula as follows;
torque, τ = F x r x sinθ
where;
- F is the applied force
- r is the perpendicular distance if force applied
F = τ /(r sinθ)
F = (39) / (0.3 sin 60)
F = 150.1 N
Thus, the magnitude (in N) of the force she must exert on the wrench is 150.1 N.
Learn more about torque here: brainly.com/question/14839816
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Answer:
Normal stress = 66/62.84 = 1.05kips/in²
shearing stress = T/2 = 0.952/2 = 0.476 kips/in²
Explanation:
A steel pipe of 12-in. outer diameter d₂ =12in d₁= 12 -4in = 8in
4 -in.-thick
angle of 25°
Axial force P = 66 kip axial force
determine the normal and shearing stresses
Normal stress б = force/area = P/A
= 66/ (П* (d₂²-d₁²)/4
=66/ (3.142* (12²-8²)/4
= 66/62.84 = 1.05kips/in²
Tangential stress T = force* cos ∅/area = P/A
= 66* cos 25/ (П* (d₂²-d₁²)/4
=59.82/ (3.142* (12²-8²)/4
= 59.82/62.84 = 0.952kips/in²
shearing stress = tangential stress /2
= T/2 = 0.952/2 = 0.476 kips/in²
