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MrRissso [65]
3 years ago
13

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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How much do you have to increase the vertical distance to double the impact speed of an object?
sergey [27]
Twice the distance to double an impact
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3 years ago
A 1500-W electric heater is plugged into the outlet of a 120-V circuit that has a 20-A circuit breaker. You plug an electric hai
Korolek [52]

Answer:

b) 900 W

Explanation:

The breaker trips when the current is equal to 20 A. The power (P) is the ddp (V) multiplied by the current. So, for the electric heater, the current is:

P = V*i

1500 = 120*i

i = 12.5 A

So, to become in 20 A, it's needed 7.5 A, which must come from the hairdryer. Its power must be:

P = 120*7.5

P = 900 W

8 0
3 years ago
A 60 kg pupil runs for 600m in 1 minute uniformly calculate kinetic energy​
AURORKA [14]

velocity = traveled distance ÷ time of the traveled distance is seconds

velocity = 600 ÷ 60

velocity = 10 m/s

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Kinetic Energy = 1/2 × mass × ( velocity )^2

KE = 1/2 × 60 × ( 10 )^2

KE = 30 × 100

KE = 3000 j

8 0
3 years ago
(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup
KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve r(\theta) = 2\cdot \sin 5\theta is 4\pi.

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx

Where

x_{o}, x_{f} - Initial and final position, respectively, measured in meters.

F(x) - Force as a function of position, measured in newtons.

Given that F = k\cdot x and the fact that F = 25\,N when x = 0.3\,m - 0.2\,m, the spring constant (k), measured in newtons per meter, is:

k = \frac{F}{x}

k = \frac{25\,N}{0.3\,m-0.2\,m}

k = 250\,\frac{N}{m}

Now, the work function is obtained:

W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx

W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]

W = 0.313\,J

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be r(\theta) = 2\cdot \sin 5\theta. The area of the region enclosed by one loop of the curve is given by the following integral:

A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta

A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta

By using trigonometrical identities, the integral is further simplified:

A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta

A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta

A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta

A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)

A = 4\pi

The area of the region enclosed by one loop of the curve r(\theta) = 2\cdot \sin 5\theta is 4\pi.

5 0
3 years ago
A particular HeNe laser beam at 633 nm has a spot size of 0.8 mm. Assuming a
Kobotan [32]

Answer:

Explanation:

A

8 0
3 years ago
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