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Sphinxa [80]
1 year ago
6

(a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500 - μC charge and flies due west at a sp

eed of 660 m/s over the Earth’s south magnetic pole, where the 8.00 10 -T 5  magnetic field points straight up. What are the direction and the magnitude of the magnetic force on the plane? (b) Discuss whether the value obtained in part (a) implies this is a significant or negligible effect.
Physics
1 answer:
12345 [234]1 year ago
8 0

(a) 2.64\cdot 10^{-8} N north

We can treat the aircraft as a single point charge moving in a magnetic field. In this case, the magnetic force exerted on the plane is

F=qvB sin \theta

where

q=0.500 \mu C = 0.500\cdot 10^{-6} C is the charge on the plane

v = 660 m/s is the velocity

B=8.00\cdot 10^{-5} T is the magnitude of the magnetic field

\theta=90^{\circ} is the angle between the direction of motion of the jet and of the magnetic field

Substituting,

F=(0.5\cdot 10^{-6})(660)(8.0\cdot 10^{-5})=2.64\cdot 10^{-8} N

The direction can be found by using Fleming's left hand rule. We have:

- index finger: magnetic field direction (straight up)

- middle finger: velocity of the plane (due west)

- force: thumb --> north

(b) Not negligible

As we can see from part (a), the magnitude of the force is not really big, so the effects are negligible.

For instance, we can compare this force with the weight of a plane. If we take a Boeing 737, its mass is about 80,000 kg, so its weight is

W=mg=(80000)(9.8)=784,000 N

As we can see, this is several orders of magnitude bigger than the magnetic force calculated at point (a), so the effects of the magnetic force are negligible.

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Show your work and resoning for the below requirement.
Leno4ka [110]

Answer:

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

Explanation:

We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.

Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn

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            - T_{y} L + W₂ L + W₁ L / 2 = 0

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we use trigonometry to find the cable tension

             sin 30 = T_{y} / T

             T = T_{y} / sin 30

             T = 195 / sin 30

             T = 390 lb

This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap

             T < 500 lb

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2 years ago
The mass of the Sun is 2 × 1030 kg, and the distance between Neptune and the Sun is 30 AU. What is the orbital period of Neptune
Veronika [31]
Kepler's third law states that, for a planet orbiting around the Sun, the ratio between the cube of the radius of the orbit and the square of the orbital period is a constant:
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where
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Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so
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And if we re-arrange the equation (1), we can find the orbital period of Neptune:
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Answer:

<u>Informed Consent- The study didn't conform this ethical standard.</u>

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