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3 years ago

Hurricane Katrina and hurricane Rita are similar in what way? (I will give brainliest to the correct answer) :)

Physics

2 answers:

scoundrel [369]3 years ago

5 0

Answer:

both same way

Explanation:

RoseWind [281]3 years ago

4 0

Answer:

Rita and Katrina both followed similar paths into the Gulf.

Explanation:

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The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de

spin [16.1K]

Answer:

the spring constant k = $5.409*10^4 \ N/m$

the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

Explanation:

From Hooke's Law

$F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m$

Thus; the spring constant k = $5.409*10^4 \ N/m$

The amplitude is decreasing 37% during one period of the motion

$e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}$

$b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) } }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s$

Therefore; the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

5 0

3 years ago

The ratio of carbon-14 to nitrogen-14 is an artifact is 1:3. Given that half-life of carbon-14 is 5730years, how old is the arti

dusya [7]

Answer:

9155 years old

Explanation:

We use the following expression for the decay of a substance:

$N = N_0\,\,e^{-k*t}$

So we first estimate the value of k knowing that the half-life of the C14 is 5730 years:

$N = N_0\,\,e^{-k*t}\\N_0/2=N_0\,\,e^{-k*5730}\\1/2 = e^{-k*5730}\\ln(1/2)=-k*5730\\k= 0.00012$

so, now we can estimate the age of the artifact by solving for"t" in the equation:

$1/3=e^{-0.00012*t}\\ln(1/3)= -0.00012*t\\t=9155. 102$

which we can round to 9155 years old.

5 0

3 years ago

A proton and an electron are fixed in space with a separation of 859 nm. Calculate the electric potential at the midpoint betwee

makkiz [27]

Answer:

The electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

Explanation:

Electric potential is given as;

V = E*r

where;

E is the electric field strength, = kq/r²

V = ( kq/r²)*r

V = kq/r

k is coulomb's constant = 8.99 X 10⁹ Nm²/C²

q is the charge of the particles = 1.6 X 10⁻¹⁹ C

r is the distance between the particles = 859 nm

At midpoint, the distance = r/2 = 859nm/2 = 429.5 nm

V = (8.99 X 10⁹ * 1.6 X 10⁻¹⁹)/ (429.5 X 10⁻⁹)

V = 3.349 X 10⁻³ Volts

Therefore, the electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

3 0

3 years ago

The acceleration reaches its minimal value of zero at the top of the trajectory. *

Serhud [2]

it is true. the trajectory reaches the value of zero at the top

4 0

3 years ago

Which two statements correctly describe the role of a semiconductor in a

KIM [24]

Answer:

It's A and C

7 0

3 years ago