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FinnZ [79.3K]
3 years ago
8

When using the right-hand rule to determine the direction of the magnetic field around a current-carrying wire, which part of th

e right hand points in the direction of the magnetic field?
palm

fingers

back

thumb
Physics
2 answers:
kirza4 [7]3 years ago
8 0

Answer:

Curled fingers will give the direction of magnetic field.

Explanation:

The right hand thumb rule states that hold any current carrying conductor in your right hand such that the thumb is pointing in the direction of electric current. Then the curled finger will give the direction of magnetic field.

For example : when the current in the wire is flowing towards right, then the curled finger will give the direction of magnetic field i.e. the field is towards bottom.

Hence, the correct option is (c) fingers.

Margarita [4]3 years ago
5 0
Your answer will be fingers

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Dafna1 [17]

Answer:

the focal length of the mirror is :  f=-2.5\,\,cm

Explanation:

Use the formula for the formation of image using a divergent mirror and recalling that the image (s') that this mirror formed is virtual, so it is entered as a negative number in the formula. Use the object position (s) as 10, the image position (s') as -2, and derive the value of the focal length:

\frac{1}{s} +\frac{1}{s'}=\frac{1}{f}\\\frac{1}{10} +\frac{1}{-2}=\frac{1}{f}\\\frac{1}{10} -\frac{1}{2}=\frac{1}{f}\\\frac{10\,f}{10} -\frac{10\,f}{2}=\frac{10\,f}{f}\\f-5\,f=10\\-4\,f=10\\f=-2.5\,\,cm

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A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel
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Answer:

1.69\cdot 10^{10}J

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

E=-G\frac{mM}{2r}

where

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M=5.98\cdot 10^{24}kg is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m

So the initial total energy is

E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J

When the satellite hits the ground, it is now on Earth's surface, so

r=R=6370 km=6.37\cdot 10^6 m

so its gravitational potential energy is

U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J

And since it hits the ground with speed

v=1.90 km/s = 1900 m/s

it also has kinetic energy:

K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J

So the total energy when the satellite hits the ground is

E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J

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\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J

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Explanation:

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