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VARVARA [1.3K]
3 years ago
9

A plane flies 1,480 miles in 4 hours, with the wind. On its return trip, it took 5 hours flying against the same wind to go 1,15

0 miles. How fast would the plane have flown without any wind? Type in your numerical answer only; do not type any words or letters with your answer.
Physics
1 answer:
Katarina [22]3 years ago
3 0

Answer:300 mi/hr

Explanation:

Given

Plane Flies 1480 miles in 4 hr with the wind

While returning it took 5 hr to cover 1150 miles against the wind

Let v and v_0 be the absolute speed of Plane and wind

Thus during with the flow of wind

time=\frac{Distance}{Speed}

\frac{1480}{v+v_0}=4

370=v+v_0   ------------1

During return trip wind speed oppose Plane thus

\frac{1150}{v-v_0}=5

230=v-v_0    ----------2

adding 1 & 2 we get

600=2v+v_0-v_0

v=300 mph

Thus Speed of plane without wind is 300 mi/hr

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The current that would pass through the 30 ohms resistor is 2 A.

<h3>What is electric current?</h3>

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To calculate the electric current that would pass through the 30 ohms resistor, we use the formula below

Formula:

  • I = V/Rt........... Equation 1

Where:

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From the question,

Given:

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Substitute these values into equation 1

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Hence, The current that would pass through the 30 ohms resistor is 2 A.

Learn more about electric current here: brainly.com/question/1100341

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The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m
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Answer:

r = 5.335 meters

Explanation:

Given that,

Charge 1, q_1=-165\ \mu C

Charge 2, q_2=115\ \mu C

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The force of attraction between two charges is given by :

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r=\sqrt{\dfrac{9\times 10^9\times 165\times 10^{-6}\times 115\times 10^{-6}}{6}}

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So, the separation between two charges is 5.335 meters. Hence, this is the required solution.

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A 50.0 g toy car is released from rest on a frictionless track with a vertical loop of radius R (loop-the-loop). The initial hei
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Answer:

the speed of the car at the top of the vertical loop  v_{top} = 2.0 \sqrt{gR \ \ }

the magnitude of the normal force acting on the car at the top of the vertical loop   F_{N} = 1.47 \ \ N

Explanation:

Using the law of conservation of energy ;

mgh = mg (2R) + \frac{1}{2}mv^2_{top}\\\\mg ( 4.00 \ R) = mg (2R) + \frac{1}{2}mv^2_{top}\\\\g(4.00 \ R) = g (2R) + \frac{1}{2}v^2 _{top}\\\\v_{top} = \sqrt{2g(4.00R - 2R)}\\\\v_{top} = \sqrt{2g(4.00-2)R

v_{top} = 2.0 \sqrt{gR \ \ }

The  magnitude of the normal force acting on the car at the top of the vertical loop can be calculated as:

F_{N} = \frac{mv^2_{top}}{R} \ - mg\\\\F_{N} = \frac{m(2.0 \sqrt{gR})^2}{R} \ - mg\\\\F_{N} = [(2.0^2-1]mg\\\\F_{N} = [(2.0)^2 -1) (50*10^{-3} \ kg)(9.8 \ m/s^2]\\\\

F_{N} = 1.47 \ \ N

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3 years ago
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