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BlackZzzverrR [31]
1 year ago
12

Which structural damage could be expected if a Category 3 hurricane is predicted to hit an area?(1 point)

Physics
1 answer:
Sphinxa [80]1 year ago
8 0

The structural damage which could be expected if a Category 3 hurricane which is predicted to hit an area is high percentage of framed homes could be destroyed, with total roof failure and wall collapse and is denoted as option C.

<h3>What is Hurricane?</h3>

This refers to the rapidly rotating storm which has a low-pressure center and is characterized by strong winds ad thunderstorms and of different types such as category 1 to 3.

Category 3 hurricane leads to loss of lives and properties which is depicted as  framed homes being destroyed with total roof failure and wall collapse which is why option C was chosen.

Read more about Hurricane here brainly.com/question/2835662

#SPJ1

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Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur
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Answer:

v_A=7667m/s\\\\v_B=7487m/s

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The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0
3 years ago
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