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Romashka-Z-Leto [24]
3 years ago
14

Install webroot mac +1 (888)(210)(2883) install webroot macbook pro

Physics
1 answer:
finlep [7]3 years ago
8 0
<h2><em><u>ookk</u></em></h2>

\huge\underline\mathcal \pink{Mr\:phenomenal}

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As a child walks, for a brief moment his entire weight is placed on one heel, which can be approximated as a small circle. Calcu
Brut [27]

The correct answer is option b, 63.6 kPa

Given,

The diameter of the heel, d=7 cm=0.07 m

The mass of the child, m=25 kg

The pressure is given by the ratio of the force to the area through which the force is applied.

The force applied on the floor by the child is equal to its weight.

Thus the pressure applied on the floor by the child is given by,

\begin{gathered} P=\frac{F}{A} \\ =\frac{mg}{\pi(\frac{d}{2})^2} \end{gathered}

Where A is the area of the cross-section of the heel.

On substituting the known values,

\begin{gathered} P=\frac{25\times9.8}{\pi(\frac{0.07}{2})^2} \\ =63.6\times10^3\text{ Pa} \\ =63.6\text{ kPa} \end{gathered}

Thus the pressure applied on the floor by the heel is 63.6 kPa

4 0
11 months ago
What is the rotational kinetic energy of the Earth about the Sun? Assume the earth is a uniform sphere, mass of the Earth is 5.9
Thepotemich [5.8K]

Answer:

2.66x10^33 J

Explanation:

In order to do this, we first need to know the expression for kinetic energy:

E = 1/2 I*w²  (1)

I is moment of innertia

w is angular speed.

Moment of Innertia can be calculated using the following expression:

I = 2/5 M*R²  (2)

M is mass of earth, R is radius of earth

Replacing the data in expression (2) we have:

I = 2/5 * 5.97x10^24 * (6.37x10^6)²

I = 9.69x10^37 kg m²

Next, we need to calculate the angular speed of Earth over it's axis, this is easy, because we know the Earth rotates over it's own axis once a day, 24 hours (86400 s), and assuming Earth is a perfect sphere, we can calculate the speed:

w = 2π / 86400 = 7.27x10^-5 rad/s

next thing we need to do is calculate the rotational kinetic energy of earth on it's axis, using equation (1) so:

E = 1/2 * 9.69x10^37 * (7.27x10^-5)²

E = 2.56x10^29 J

Now that we have this value, we can finally calculate the rotational kinetic energy of earth about the sun. For that, we need to calculate again the angular speed of earth about the sun. The Earth rotates around the sun once a year, or 365 days, which is 3.1536x10^7 s, so the angular speed would be:

w = 2π/3.1536x10^7 = 1.99x10^-7 rad/s

finally the energy is the combination of the sun and earth so:

K = 1/2 (Ie + Me*Rorb²)wo²

I is innertia for earth

Me mass of earth

Rorb RAdius of orbit around the sun

wo is angular speed around the sun

Replacing the data we finally have:

K = 1/2 [9.69x10^37 + 5.97x10^24 * (1.5x10^11)²]*(1.99x10^-7)²

K = 2.66x10^33 J

4 0
3 years ago
Need some help please answer please
yuradex [85]

Answer:

Its not d or e

Explanation:

8 0
2 years ago
What will happen to the speed of an object if the net force is in the direction of the motion?
nata0808 [166]
The speed increases
6 0
3 years ago
Read 2 more answers
The type of bond formed by shared electrons is covalent, ionic, or metallic
loris [4]
I believe the answer in Covalent Bond.
7 0
3 years ago
Read 2 more answers
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