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Goshia [24]
3 years ago
8

In 1986, the first flight around the globe without a single refueling was completed. The aircraft’s average speed was 186 km/h.

If the airplane landed at this speed and accelerated at −1.5 m/s2 , how long did it take for the airplane to stop?
Physics
1 answer:
Whitepunk [10]3 years ago
3 0

Answer: 34.4 sec

Explanation: Assuming that the desacceleration was constant, we have  a = -1.5m/s²

vi = 186 km/h → 51.6 m/s

vf = 0 (because it stopped)

Vf = Vi + at

0 = 51.6 - 1.5t

1.5t = 51.6

t = 34.4 s

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the question is correct

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How can the strength of an electromagnet be increased?
rewona [7]

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You can make an electromagnet stronger by doing these things: wrapping the coil around a piece of iron (such as an iron nail) adding more turns to the coil. increasing the current flowing through the coil.

Explanation:

5 0
2 years ago
A harvest mouse can detect sounds below the threshold of human hearing, as quiet as −10 dB. Suppose you are sitting in a field o
PtichkaEL [24]

Answer:

The distance from harvest mouse to the leaf is 4.74 m.

Explanation:

Given that,

Intensity = -10 dB

Distance = 1.5 m

We need to calculate the power of intensity

Using formula of power of intensity

P=I_{0}A

P=I_{0}\times4\pi r^2

Put the value into the formula

P=1.0\times10^{-12}\times4\pi\times(1.5)^2

P=28.27\times10^{-12}\ W/m^2

We need to calculate the minimum intensity level

Using formula of minimum intensity

I=10\log_{10}(\dfrac{I}{I_{0}})

-10=10\log_{10}(\dfrac{I}{I_{0}})

\log_{10}\dfrac{I}{I_{0}}=-1

I=10^{-1}\times(1.0\times10^{-12})

I=1\times10^{-13}\ W/m

We need to calculate the area

Using formula of intensity

I=\dfrac{P}{A}

A=\dfrac{P}{I}

Put the value into the formula

A=\dfrac{28.27\times10^{-12}}{1\times10^{-13}}

A=282.7\ m^2

We need to calculate the distance

Using formula of area

A=4\pi r^2

Put the value into the formula

282.7=4\pi\times r^2

r^2=\dfrac{282.7}{4\pi}

r=\sqrt{\dfrac{282.7}{4\pi}}

r=4.74\ m

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6 0
2 years ago
A proton is shot perpendicularly at an infinite plane of charge. The charge density of the plane is +7.65×10^−4 C/m^2. If the pr
Alexeev081 [22]

Answer:

The velocity is 2.94\times10^{6}\ m/s.

Explanation:

Given that,

Charge density of the plane \sigma=7.65\times10^{−4}\ C/m^2

Distance = 1.05 mm

We need to calculate the electric field due to plane of charge

Using formula of electric field

E=\dfrac{\sigma}{2\epsilon}

Put the value into the formula

E=\dfrac{7.65\times10^{−4}}{2\times8.85\times10^{-12}}

E=4.322\times10^{7}\ N/C

We need to calculate potential difference

Using formula of potential difference

V=E\times r

Put the value into the formula

V=4.322\times10^{7}\times1.05\times10^{-3}

V=4.5381\times10^{4}\ Volt

We need to calculate the work requires to be done to reach the surface of the plane

Using formula of work done

W=qV

Put the value into the formula

W = 1.6\times10^{-19}\times4.5381\times10^{4}

W=7.26096\times10^{-15}\ J

We need to calculate the velocity

Using work energy theorem

W=\dfrac{1}{2}mv^2

v^2=\dfrac{2W}{m}

v=\sqrt{\dfrac{2\times7.26096\times10^{-15}}{1.67\times10^{-27}}}

v=2.94\times10^{6}\ m/s

Hence, The velocity is 2.94\times10^{6}\ m/s.

8 0
2 years ago
Please help ASAP! Thank you very much!
Nitella [24]

Answer:D

Explanation:

Because you ether use addition o multiplication to find your answer

7 0
2 years ago
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