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Oduvanchick
2 months ago
10

Honey bees can acquire a small net charge on the order of 1 pC as they fly through the air and interact with plants. Estimate th

e magnetic force on a honey bee due to Earth's magnetic field as the bee flies near the ground from east to west. Average speed of a bee is 15 miles/hour.
HINT: Model the honey bee as a moving, charged point particle. The direction the bee flies is significant.

FEEDBACK: Model the charged honey bee as a point particle.
Physics
1 answer:
Softa [3K]2 months ago
7 0
The force acting upon a charged particle in the presence of a magnetic field can be described by the equation: where q symbolizes the particle's charge, v represents its velocity, B indicates the magnetic field strength, and θ is the angle between the vectors of B and v. In this context, we consider: q as the charge of a honey bee; v as the flying speed of the bee; B as the Earth's magnetic field's average strength; noting that the bee's motion from east to west contrasts with the south to north direction of the magnetic field. By substituting these parameters into the equation, we arrive at an estimate.
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An 8.0 m, 240 N uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground
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Answer:

5.7 m

Explanation:

AD = length of the ladder = L = 8 m

AB = the position of the ladder's center of mass = (0.5) L = (0.5) 8 = 4 m

AC = distance of the climber from the bottom of the ladder = x

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F_{g} = weight of the climber = 710 N

F = force exerted by the wall on the ladder

N = normal force acting on the ladder from the ground =?

By applying force equilibrium in the vertical direction

N = F_{g} + W

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μ = Coefficient of static friction = 0.55

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Static friction force can be expressed as

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f = (0.55) (950)

f = 522.5 N

The equation for force along the horizontal axis reads

F = f

F = 522.5 N

using torque equilibrium around point A

F Sin50 (AD) = W Cos50 (AB) + (F_{g} Cos50 (AC))

(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)

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