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Elanso [62]
3 years ago
3

Why are hurricanes considered more damaging than tornadoes when tornados have stronger winds? a. Hurricanes only strike coastlin

es while tornados strike coasts and inland. b. Hurricanes also cause flooding and occur over a broader area. c. Tornadoes move more slowly than hurricanes. d. Tornadoes only occur inland, away from coastlines.
Physics
2 answers:
aalyn [17]3 years ago
6 0
Well,

We call something damaging when it affects us humans.  In that sense, then, hurricanes that occur far away from the coastlines would not be considered "damaging."

Option A would not support the argument.
Option D would not support the argument either.

We know that hurricanes cause flooding (because they occur near the coastlines) and that they are (hundreds of times) bigger than tornadoes, so Option B is correct.
viktelen [127]3 years ago
3 0

A tornado lasts a few hours, and affects people and things in an area that's
a few hundred yards wide and maybe a few tens of miles long.

A hurricane lasts <em><u>weeks</u></em>, and can affect people and things in an area that's
a few hundred <em><u>miles</u></em> across and a few <em><u>thousand</u></em> miles long.

The hurricane is the tortoise to the tornado's hare.


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8 0
3 years ago
Read 2 more answers
Problem 4 A meteoroid is first observed approaching the earth when it is 402,000 km from the center of the earth with a true ano
Nikitich [7]

Answer:

Part a: The eccentricity is 1.086.

Part b: The altitude at closest approach is 5088 km

Part c: The velocity at perigee is 8.516 km/s

Part d: The turn angle is 134.08 while the aiming radius is 5641.28 km

Explanation:

<h2>Part a </h2>

Specific energy is given by

\epsilon=\frac{v^2}{2}-\frac{\mu}{r}

Here

  • ε is the specific energy
  • v is the velocity which is given as 2.23 km/s
  • μ is the gravitational constant whose value is 398600
  • r is the distance between earth and the meteorite which is 402,000 km

                         \epsilon=\frac{v^2}{2}-\frac{\mu}{r}\\\epsilon=\frac{2.2^2}{2}-\frac{398600}{402,000}\\\epsilon=1.495 km^2/s^2

Value of specific energy is also given as

\epsilon=\frac{\mu}{2a}\\a=\frac{\mu}{2\epsilon}\\a=\frac{398600}{2\times 1.495}\\a=13319 km

Orbit formula is given as

r=a(\frac{e^2-1}{1+ecos \theta})\\ae^2-recos\theta-(a+r)=0

Putting values in this equation and solving for e via the quadratic formula gives

ae^2-recos\theta-(a+r)=0\\(133319)e^2-(402000)(cos 150) e-(133319+402000)=0\\133319e^2+348142.21 e-535319=0\\\\e=\frac{-348142.21 \pm \sqrt{348142.21^2-4(133319)(535319)}}{2 (133319)}\\\\e=1.086 \, or \, -3.69

As the value of eccentricity cannot be negative so the eccentricity is 1.086.

<h2>Part b</h2>

The radius of trajectory at perigee is given as

r_p=a(e-1)\\

Substituting values gives

r_p=133319 (1.086-1)\\r_p=11465.4 km

Now for estimation of altitude z above earth is given as

z=r_p-R_E\\z=11465.4-6378\\z=5087.434\\z\approx 5088 km

So the altitude at closest approach is 5088 km

<h2>Part c</h2>

radius of perigee is also given as

r_p=\frac{h^2}{\mu}\frac{1}{1+e}

Rearranging this equation gives

h=\sqrt{r_p\mu(1+e)}\\h=\sqrt{11465.4 \times 3986000 \times (1+1.086)}\\h=97638.489 km^2/s

Now the velocity at perigee is given as

v_p=\frac{h}{r_p}\\v_p=\frac{97638.489}{11465.4}\\v_p=8.516 km/s\\

So the velocity at perigee is 8.516 km/s

<h2>Part d</h2>

Turn angle is given as

\delta =2 sin^{-1} (\frac{1}{e})

Substituting value in the equation gives

\delta =2 sin^{-1} (\frac{1}{e})\\\delta =2 sin^{-1} (\frac{1}{1.086})\\\delta =134.08

Aiming radius is given as

\Delta =a \sqrt{e^2-1}

Substituting value in the equation gives

\Delta =a \sqrt{e^2-1}\\\Delta =13319 \sqrt{1.086^2-1}\\\Delta=5641.28 km

So the turn angle is 134.08 while the aiming radius is 5641.28 km

3 0
4 years ago
Projectile A is launched horizontally at a speed of 20 meters per second from the top of a cliff and strikes a level surface bel
pogonyaev

Answer: 3 seconds

Explanation:

Initial velocity(u) of projectile A in vertical direction = 0m/s

acceleration due to gravity a=g=9.81m/s^2

Time taken(t) of projectile A = 3s

Initial velocity of projectile B = 0m/s(vertical direction)

We can get height of cliff using parameters of projectile A since it's the same location.

Height(S) = u×t + 0.5×a×t^2

u =0

S= 0.5×9.81×3^2 = 44.145m

Time taken for projectile B to reach the ground:

S = u×t + 0.5×a×t^2

u =0, S=44.145m, a=9.81m/s^2

44.145 = 0.5×9.81×t^2

44.145 = 4.905×t^2

44.145 ÷ 4.905 = t^2

9 = t^2

t = sqrt(9)

t = 3seconds

5 0
4 years ago
A long solid conducting cylinder with radius a = 12 cm carries current I1 = 5 A going into the page. This current is distributed
jenyasd209 [6]

Answer:

Explanation:

We shall use Ampere's circuital law to find magnetic field at required point.

The point is outside the circumference of two given wires so whole current will be accounted for .

Ampere's circuital law

B = ∫ Bdl = μ₀ I

line integral will be over circular path of radius r = 41 cm .

Total current  I  = 5A -3A = 2A .

∫ Bdl = μ₀ I

2π r B = μ₀ I

2π x .41  B = 4π x 10⁻⁷ x 2

B = 2 x 10⁻⁷ x 2 / .41

= 9.75 x 10⁻⁷ T . It will be along - ve Y - direction.

7 0
3 years ago
Karen travels 10 mile in 1 hour. How many
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I think the answer is 120 minutes hope this helps
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