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3 years ago

How did researchers go about investigating trans-sonic flight

1 answer:

6 0

<span>A wind tunnel is a traditional tool used for trans-sonic flight research. But, during those days, it’s not exact and consistent to use for aircraft flow conditions. The reason is that wind tunnel models will be visible to airflow that produces shock waves. So the U.S and Britain develop an instrument to get data and record flow conditions for a specified speed, these are the specialized research airplanes.</span>

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Three magnets are placed on a plastic stick as shown in the image. Explain how the magnets need to be rearranged so that they st

dimaraw [331]

Answer:

The magnets need to be placed with red closest to blue.

Opposite poles attract.

The magnets will be attracted to each other with enough force to stick together.

Explanation:

6 0

3 years ago

Read 2 more answers

How far will an object move in 6 s if its average

Dafna11 [192]

Given:-

Time taken by the particle (t) = 6 s
Average speed (v) = 40 m/s

To Find: Distance (s) travelled by the particle.

We know,

s = vt

where,

s = Distance travelled,
v = Speed &
t = Time taken.

Putting the values,

s = (40 m/s)(6 s)

→ s = 240 m ...(Ans.)

6 0

3 years ago

The field inside a charged parallel-plate capacitor is __________.

Bess [88]

<span>Uniform. A parallel plate capacitor is a simple arrangement of electrodes and dielectric to form a capacitor. Here two parallel conductive plates are used as electrodes with a medium or dielectric in between them. Charge separation in a parallel-plate capacitor causes an internal electric field, which is uniform.</span>

5 0

4 years ago

Read 2 more answers

The drinking water needs of an office are met by cooling tab water in a refrigerated water fountain from 23 to 6°C at an average

Ratling [72]

Answer:

The correct option is;

Explanation:

Here we have the Coefficient Of Performance, COP given by

$COP = \frac{Q_{cold}}{W} = 3.1$

The heat change from 23° to 6°C for a mass of 10 kg/h which is equivalent to 10/(60×60) kg/s or 2.78 g/s we have

$Q_{cold}$ = m·c·ΔT = 2.78 × 4.18 × (23 - 6) = 197.39 J

Therefore, plugging in the value for $Q_{cold}$ in the COP equation we get;

$COP = \frac{197.39 }{W} = 3.1$ which gives

$W = \frac{197.39 }{3.1} = 63.674 \ J$

Since we were working with mass flow rate then the power input is the same as the work done per second and the power input to the refrigerator = 63.674 J/s ≈ 64 W.

The power input to the refrigerator is approximately 64 W.

4 0

3 years ago

Read 2 more answers

A lamp is connected to the power supply.

Vika [28.1K]

Answer:

There are 45 turns in the secondary coil.

Explanation:

Given that,

Input potential of the lamp, $V_{in}=5\ V$