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liubo4ka [24]
3 years ago
12

How long does it take an airplane to fly 5000 miles if it maintains a speed of 240 miles per hour

Physics
1 answer:
aliina [53]3 years ago
3 0
With a 30 mph head wind it takes the plane 18.52 hours to fly 5000 miles. ANSWER 2: With a 30 mph tail wind it takes the plane 15.15 hours to fly 5000 miles.
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In places such as hospital operating rooms or factories for electronic circuit boards, electric sparks must be avoided. A person
FrozenT [24]

Time taken by the rubber-soled shoes to reduce a person's potential from 3.00×10³ V to 100V is 3.91s

The initial potential difference \Delta V_o is related to the final potential difference \Delta V by

\Delta V=\Delta V_{o} e^{-t / \tau}

Our target is to find t, so we rearrange equation (1) for t to be

\Delta V=\Delta V_{o} e^{-t / RC}\\ \ln \left ( \frac{\Delta V}{\Delta V_{o}} \right ) = \frac{-t}{RC}\\ t= -RC \ln \left ( \frac{\Delta V}{\Delta V_{o}} \right ) \tag{2}

The body has a capacitance 150 pF while the foot has a capacitance 80 pF and both are in parallel connection. So, the equivalent capacitance is

C= 150 \mathrm{~pF} + 80 \mathrm{~pF} = 230 \mathrm{~pF}

The shoes of the rubber-soled has resistance R=5000 \mathrm{~M\Omega}. Plug the values for \Delta V, \Delta V_o and C into equation (2) to get the time t.

t&= -RC \ln \left ( \frac{\Delta V}{\Delta V_{o}} \right ) \\ &= -( 5000 \times 10^{6} \mathrm{~\Omega})(230 \times 10^{-12} \mathrm{~F}) \ln \left ( \frac{100\mathrm{~V}}{3000 \mathrm{~V}} \right ) \\ &= {3.91 \mathrm{~s}}

Learn more about capacitance here:

brainly.com/question/12644355

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2 years ago
Explain how you know that gasoline burning in a car engine is a chemical change.?
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Burning, in general, is a chemical change.
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5 0
3 years ago
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A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal
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Answer:

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The tension of the string does no work because the tension does not

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3 years ago
First Amendment
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Answer:

They were concerned about the government having too much power.

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An upward force act on a proton as it moves with a speed of 2.0 x 10^5 meters/seconds through a magnetic field of 8.5 x 10^2
Gekata [30.6K]

Force on a moving charge is given by formula

\vec F = q(\vec v \times \vec B)

here we know that this force will be maximum when velocity is perpendicular to magnetic field

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here we know that

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now we have

F = (1.6 \times 10^{-19})(2 \times 10^5)(8.5 \times 10^2)

F = 2.72 \times 10^{-11} N

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