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fiasKO
3 months ago
14

Which trailer has more downward pressure where it attaches to the car?

Physics
1 answer:
serg [3.5K]3 months ago
4 0

The trailer that is loaded the most. The total weight does not matter; it is about how the load is distributed. For example, our 12,000 lb snow cat trailer has weight distribution that results in less than 100 lbs of tongue weight. Heavy tongue weight can create issues, as it can shift the weight off the front wheels of the towing vehicle, causing instability.

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A beam of electrons moves at right angles to a magnetic field of 4.5 × 10-2 tesla. If the electrons have a velocity of 6.5 × 106
Softa [3030]
Greetings!

Using the formula F = Bqv sin theta, we define F as Force, B as magnetic flux density, q as charge, v as velocity, and theta as the angle created by the moving electrons in relation to the magnetic field.

^^^You can compute the force using that equation^^^

In conclusion, your result would MOST LIKELY be "B".

"<span>-3.9 × 10-14 N"
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<span>I trust my response has been beneficial. Thank you for your question. We look forward to assisting with more inquiries. Have a wonderful day ahead!:</span>
6 0
2 months ago
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An electric buzzer is activated, then sealed inside a glass chamber. When all of the air is pumped out of the chamber, how is th
ValentinkaMS [3465]
The intensity of the sound increases because sound waves are mechanical waves, meaning they cannot move through a vacuum and require a medium to propagate.
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2 months ago
How do electromagnetism and gravitation differ from the strong and weak nuclear forces?
ValentinkaMS [3465]
Electromagnetism, which deals with interactions between charged particles, has a considerable range and can create forces that repel like charges while attracting opposites. In contrast, the weak nuclear force is very short-ranged and isn't classified as a force.
5 0
1 month ago
As computer structures get smaller and smaller, quantum rules start to create difficulties. Suppose electrons move through a cha
Softa [3030]

Answer:

a) ∆x∆v = 5.78*10^-5

∆v = 1157.08 m/s

b) 4.32*10^{-11}

Explanation:

This problem can be addressed using Heisenberg's uncertainty principle, which is expressed as:

\Delta x\Delta p \geq \frac{\hbar}{2}

Where h represents Planck’s constant (6.62*10^-34 J s).

Assuming that the electron's mass remains the same, we proceed as follows:

\Delta x \Delta (m_ev)=m_e\Delta x\Delta v \geq \frac{\hbar}{2}

Utilizing the electron's mass (9.61*10^-31 kg) and the uncertainty in position (50 nm), we can compute ∆x∆v and ∆v:

\Delta x \Delta v\geq\frac{\hbar}{2m_e}=\frac{(1.055*10^{-34}Js)}{2(9.1*10^{-31}kg)}=5.78*10^{-5}\ m^2/s

\Delta v\geq\frac{5.78*10^{-5}}{50*10^{-9}m}=1157.08\frac{m}{s}

If we treat the electron like a classic particle, the time required to cross the channel is determined using the upper limit of the uncertainty in velocity:

t=\frac{x}{v}=\frac{50*10^{-9}m}{1157.08m/s}=4.32*10^{-11}s

6 0
3 months ago
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