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gogolik [260]
3 years ago
14

Assume the average value of the vertical component of Earth's magnetic field is 42 μT (downward) in some region that has an area

of 3.71 × 105km2. What then is the magnitude of the net magnetic flux through the rest of Earth's surface (the entire surface excluding that region).
Physics
1 answer:
Oliga [24]3 years ago
3 0

Answer:

The net magnetic flux through the rest of Earth's surface is  5.782\times10^{-2}\ T

Explanation:

Given that,

Magnetic field = 42 μT

Area A=3.71\times10^{5}\ km^2

A=3.71\times10^{11}\ m^2

We need to calculate the flux per unit area

flux\ per\ unit\ area=\dfrac{42\times10^{-6}}{3.71\times10^{11}}

flux\ per\ unit\ area=1.132\times10^{-16}\ T/m^2

We need to calculate the total earth's surface area

A'=4\pi r^2

A'=4\times\pi\times(6.3781\times10^{6})^2

A'=5.1120\times10^{14}\ m^2

We need to calculate the rest of earth's area

A''=A-A'

Put the value into the formula

A''=5.1120\times10^{14}-3.71\times10^{11}

A''=5.10829\times10^{14}\ m^2

We need to calculate the net magnetic flux through the rest of Earth's surface

B'=5.10829\times10^{14}\times1.132\times10^{-16}

B'=5.782\times10^{-2}\ T

Hence, The net magnetic flux through the rest of Earth's surface is  5.782\times10^{-2}\ T

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lana66690 [7]
According to Newton's Second Law of Motion, the net force experienced by the system is equal to the mass of the system in question times the acceleration in motion. In this case, the net force is the difference of gravitational force and the force experience by the motion of the airplane. This difference is already given to be 210 N.

Net force = ma
210 N = (73 kg)(a)
a = +2.92 m/s²

Thus, the acceleration of the airplane's motion is 2.92 m/s² to the positive direction which is upwards.
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An electric heater rated 600 w operates 6 hours per day find the cast to operate it for 30 days , at rs. 4.00 per unit​
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Rs. 432*10^3 (In kilowatts per hour)

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The fast train known as the TGV (Train à Grande Vitesse) that runs south from Paris, France, has a scheduled average speed of 21
andrezito [222]

Answer:

Explanation:

Given

average speed of train(v_{avg})=216 kmph\approx 60 m/s

Maximum acceleration=0.05g

Now centripetal acceleration is

a_c=\frac{v^2}{r}

0.05\times 9.8=\frac{60^2}{r}

r=7346.93 m

(b)Radius of curvature=900 m

therefore a_c=\frac{v^2}{r}

v=\sqrt{a_cr}

v=\sqrt{0.05\times 9.8\times 900}

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8 0
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A 12kg cheetah accelerates 24 m/s". What is the force the cheetah needed to run?
Kobotan [32]

Answer:

288N

Explanation:

Given parameters:

Mass of Cheetah = 12kg

Acceleration  = 24m/s²

Unknown:

Force needed by the cheetah to run  = ?

Solution:

The force needed by the Cheetah to run is the net force.

According to Newton's law;

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Insert the given parameters and solve;

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7 0
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Calculate the force of gravity between a comet with a mass of 500kg and a small asteroid with a mass of 20kg that is separated b
givi [52]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

The equivalent gravitational force is ~

  • F  \approx1.48\times 10 {}^{ - 7}  \: \: N

\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}

We know that ~

\huge\boxed{\mathrm{F = \dfrac{ Gm_1m_2}{ r²}}}

where,

  • F = gravitational force

  • m_1 = mass of 1st object = 500 kg

  • m_2 = mass of 2nd object = 20kg

  • G = gravitational constant = 6.674 × {10}^ {-11}

  • r = distance between the objects = 2.12 m

Let's calculate the force ~

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  • F = \dfrac{6.674  \times 10 {}^{ - 11} \times 10 {}^{4} }{4.4944}

  • F =  \dfrac{6.674}{4.4944}  \times 10 {}^{ - 7}

  • F =1.484 \times 10 {}^{ - 7}  \: \: newtons
7 0
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