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padilas
12 days ago
11

How is earth outer layer different from a cracked hard-boiled egg?

Physics
1 answer:
Softa [3K]12 days ago
7 0

Answer:

The surface of the Earth supports living organisms and features intricate layers within it (also, it's not just white) and, naturally, there are no oceans.

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A 68 kg hiker walks at 5.0 km/h up a 9% slope. The indicated incline is the ratio of the vertical distance and the horizontal di
Softa [3030]

The formula used to calculate power is:

Power = Work / Time

Where Work is defined as Force multiplied by Distance, hence:

Power = Force * Distance / Time

This can also be expressed as:

Power = Force * Velocity<span>


Converting the velocity from km/h to m/s:</span>

Velocity = (5 km / h) (1000 m / km) (1 h / 3600 s)<span>
Velocity = 1.39 m/s 

</span>

The calculated force corresponds to the vertical component of the hiker's weight:

Force = Wy * g = W * sin θ * g

 

To find the angle θ, we utilize the slope, which is defined as:

slope = tan θ = ratio of vertical to horizontal distance

tan θ = 0.09

θ = 5.14˚

Thus, we have Force = 68 kg * sin(5.14˚) * 9.8 m/s²<span>
Force = 666.4 * sin(5.14) = 59.73 N 

</span>

<span>Now, let's compute the power:
<span>Power = 59.73 N * 1.39 m/s = 82.96 W
Given the hiker's efficiency is at 25%, we can find the metabolic power: 
<span>Metabolic Power = 82.96 W / 0.25 = 331.83 watts </span></span></span>
6 0
1 month ago
A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.
Sav [3153]

Response:

A=0.199

Clarification:

We know that  

Mass of spring=m=450 g==\frac{450}{1000}=0.45 kg

Where 1 kg=1000 g

Frequency of oscillation=

\nu=1.2Hz

Energy for oscillation is 0.51 J

To determine the amplitude of oscillations.

Energy for oscillator=E=\frac{1}{2}m\omega^2A^2

Where \omega=2\pi\nu=Angular frequency

A=Amplitude

\pi=\frac{22}{7}

Using the formula

0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2

A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398

A=\sqrt{0.0398}=0.199

Therefore, the amplitude of oscillation=A=0.199

4 0
25 days ago
Evaluate the final kinetic energy of the supply spacecraft for the actual tractor beam force, F(x)=αx3+βF(x)=αx3+β.
Sav [3153]

Answer:

K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x

Explanation:

To calculate the kinetic energy variation, we can utilize the work-energy theorem.

W = ΔK

∫ F .dx = K - K₀

If the object starts from rest, then K₀ = 0.

So, ∫ F dx cos θ = K.

As the force and displacement directions align, the angle is zero, and hence the cosine is 1.

Now we can substitute and perform integration:

α ∫ x³ dx + β ∫ dx = K.

Thus, α x⁴ / 4 + β x = K.

Next, we evaluate from the limits F = 0 to F:

α (x⁴ / 4 - 0) + β (x - 0) = K.

Consequently, K = αX⁴ / 4 + β x.

This results in K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x.

To finalize the computation, we need to ascertain the displacement.

8 0
1 month ago
Calculate the average charge on arginine when ph=9.20. (hint : find the average charge for each ionizable group and sum these to
Sav [3153]
Arginine is classified as a basic amino acid since it has two amino groups alongside a single acid group. At a low pH level, all ionizable groups are protonated. As the pH rises slightly, the acid group loses its proton. When the pH increases further, one of the amino groups also loses a proton. At considerably high pH levels, none of the ionizable groups remain protonated.

Pkas

<span> <span><span> <span> pka1 = 1.82 </span> <span> pka2 = 8.99 </span> <span> pka3 = 12.48 </span> </span> </span></span> Thus, 9.20 is above the second pKa and below the third pKa. This indicates that the acid has already lost its proton, as has one of the amino groups, while the second amino group remains protonated. When an acid is not protonated, it carries a negative charge. An unprotonated amino group is neutral, whereas when protonated, the amino group bears a positive charge. Therefore, this amino acid exhibits one positive charge (from one of the amino groups) and one negative charge (from the acid), resulting in an overall neutral charge.
4 0
15 days ago
Using evidence from the article, defend the concept that Earth’s magnetic poles have swapped places over time.
Maru [3345]

Answer: Research has revealed signs of the Earth's magnetic field flipping in ocean floor rocks, especially at tectonic plate boundaries. These rocks exhibit alternating polarity because of the magnetization that happened as they solidified. Through radio metric dating, scientists suggest these reversals take place roughly every few hundred thousand years.

Explanation:

6 0
21 day ago
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