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dexar [7]
2 years ago
6

A small, single engine airplane is about to take off. The airplane becomes airborne, when its speed reaches 193.0 km/h. The cond

itions at the airport are ideal, there is no wind. When the engine is running at its full power, the acceleration of the airplane is 2.80 m/s2. What is the minimum required length of the runway
Physics
1 answer:
siniylev [52]2 years ago
4 0

Answer:

513 m

Explanation:

We have;

final speed of the airplane = 193.0 km/h * 1000/3600 = 53.6 m/s

acceleration of the air plane = 2.80 m/s2

initial velocity of the airplane = 0 m/s

length of the runway = distance covered

v^2 = u^2 + 2as

v^2 - u^2 = 2as

s = v^2 - u^2/2a

s = (53.6)^2 - 0^2/ 2 *  2.80

s = 2872.96/ 5.6

s = 513 m

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Answer:

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Explanation:

correct me if I'm wrong

4 0
2 years ago
A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an
olya-2409 [2.1K]

Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = \pi r^2

V = Potential applied = 2 mV

k = Dielectric constant = 40

\epsilon_0 = Electric constant = 8.854\times 10^{-12}\ \text{F/m}

Capacitance is given by

C=\dfrac{k\epsilon_0A}{d}

Charge is given by

Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}

Number of electron is given by

n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}

The number of charge carriers that will accumulate on this capacitor is approximately 289282.

6 0
2 years ago
In a thunder and lightning storm there is a rule of thumb that many people follow. After seeing the lightning, count seconds to
lubasha [3.4K]

Answer:

2.837% less than actual value.

Explanation:

Based on given information let's calculate our value.

S = Vxt = 331m/s x 5s = 1655m, that is the total distance that sound would travel in 5 seconds.

1mile = 1609.34meters.

percentage error is.

\frac{actual-calculated}{actual} *100 = \frac{1609.34-1655}{1609.34} *100 = -2.83%

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3 0
3 years ago
When you observe a physical property the substance
Amanda [17]
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5 0
3 years ago
Read 2 more answers
Church organs have a set of pipes with different lengths. With those different pipes organs can produce sounds over a wide range
Paraphin [41]

Answer: The shortest possible wavelength of sound the organ can produce is 0.224 m

Explanation:

To calculate the wavelength of light, we use the equation:

\lambda=\frac{c}{\nu}

where,

\lambda = wavelength of the light

c = speed of light = 331m/s

As wavelength and frequency follows inverse relation, shortest wavelength is produced by highest frequency.

\nu = highest frequency of light = 1.48kHz=1.48\times 10^3Hz=1.48\times 10^3s^{-1}       1Hz=1s^{-1}

\lambda=\frac{331m/s}{1.48\times 10^3s^{-1}}=0.224m

Thus the shortest possible wavelength of sound the organ can produce is 0.224 m

8 0
3 years ago
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