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cluponka [151]
3 years ago
7

A hurricane is a huge storm that forms over warm ocean waters. Many hurricanes enter the Gulf of Mexico and move in a northweste

rly direction. A hurricane travels a distance of 20 miles in 1 hour. Determine the hurricane's speed. (SPEED = DISTANCE / TIME)
Physics
1 answer:
labwork [276]3 years ago
5 0

Answer:

536.448m/s or 20miles/hr

Explanation:

Speed is defined as the distance travelled by a body or an object in a specified time.

Speed = Distance/Time

If a hurricane travels a distance of 20 miles in 1 hour.

Distance travelled = 20miles

Time taken = 1hour

Speed = 20miles/1hour

Speed = 20miles/hr

Since 1mile = 1609.344m

20miles = 20×1609.344

20miles = 32186.88m

1hour = 60seconds

Expressing the speed in m/s

Speed = 32186.88m/60s

Speed = 536.448m/s

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A uniformly charged rod of length L = 1.3 m lies along the x-axis with its right end at the origin. The rod has a total charge o
Mars2501 [29]

Answer:

E = 3544.44 N/C

Explanation:

Given:

- charge Q = 2.2 *10^-6 C

- Length L = 1.3 m

Find:

The Electric Field strength E @ a  = 1.8 m

Solution:

- The differential electric field dE due to infinitesimal charge dq can be considered as a point charge at a distance of r is given by:

                                  dE = k*dq / r^2

- The charge Q is spread over entire length L, hence:

                                  dq = (Q / L ) * dx

-The resulting dE:

                                 dE = (k*Q/L)*(dx / r^2)

- point P lies on the x- axis with distance (x+a) from differential charge from:

                                 dE = (k*Q/L)*(dx / (x+a)^2)  

- Integrate dE over length 0 to L

                                 E = (-k*Q/L)*( 1 / (x+a) )

                                 E =  (-k*Q/L)* (1 / a - 1 / (L+a))

                                 E =  (-k*Q/L)* (L / a(L+a))

                                 E = (k*Q / a(L+a))

- Evaluate E @ a = 1.8 m

                                 E =(8.99*10^9 * 2.2*10^-6 / 1.8*(1.3+1.8))

                                 E = 3544.44 N/C

8 0
3 years ago
The gage pressure of an automobile tire is measured to be 210kPa before a trip and 220kPa after the trip at a location where the
Katarina [22]

Answer:

T_{f} = 312.348\,K\,(39.198\,K)

Explanation:

Let assume that gas inside the automobile tire behaves as an ideal gas. Due to the absence of leakages, the number of moles remains constant during the trip. Air temperature can be found by using the following relation:

\frac{P_{o}}{T_{o}}=\frac{P_{f}}{T_{f}}

Final temperature is cleared with the expression:

T_{f} = \frac{P_{f}}{P_{o}}\cdot T_{o}

T_{f} = \frac{220\,kPa}{210\,kPa}\cdot (298.15\,K)

T_{f} = 312.348\,K\,(39.198\,K)

7 0
3 years ago
neptune is an average distance of 4.5×10^12m from the sun. Estimate the length of the Neptunian year.
Vikentia [17]

As per Kepler's third law we know that

\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}

now here we know that

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R_2 = Distance of Earth from Sun

so now we will have

\frac{T_1^2}{1} = \frac{(4.5 \times 10^{12})^3}{(1.5 \times 10^11)^3}

T_1^2 = 27000

T_1 = 164.3 years

so length of year of Neptune is 164.3 years

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Answer:

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Explanation:

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