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alukav5142 [94]
3 years ago
5

A very humble bumble bee is flying horizontally due North at a constant speed of 3.11 m/s. At the current location of the bumble

bee the magnetic field of Earth is 1.05×10-5 T and it points 35.1° below the horizontal. The bumble bee carries a positive electric charge of 22.5 nC. What is the size of the magnetic force acting on the bumble bee?
Physics
1 answer:
Reil [10]3 years ago
3 0

To solve this problem we will apply the concepts of the Magnetic Force. This expression will be expressed in both the vector and the scalar ways. Through this second we can directly use the presented values and replace them to obtain the value of the magnitude. Mathematically this can be described as,

\vec{F_B} = q(\vec{V}\times \vec{B})

F_B = q|v||B| sin\theta

Here,

q = Charge

v = Velocity

B = Magnetic field

\theta = \text{Angle between } \vec{B} \text{ and } \vec{V}

Our values are given as,

\theta = 35.7\°

q = 22.5*10^{-9}C

B = 1.05*10^{-5}T

v = 3.11m/s

Replacing,

F_B = (22.5*10^{-9}C)(3.11 \times 1.05*10^{-5}) sin(35.1\°)

F_B = 4.224*10^{-13}N

Therefore the size of the magnetic force acting on the bumble bee is 4.22*10^{-13}N

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Answer:

car B will be 30 Km ahead of car A.

Explanation:

We'll begin by calculating the distance travelled by each car. This is illustrated below:

For car A:

Speed = 40 km/h

Time = 3 hours

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Finally, we shall determine the distance between car B an car A. This can be obtained as follow:

Distance travelled by car B (D₆) = 150 Km

Distance travelled by car A (Dₐ) = 120 Km

Distance apart =?

Distance apart = D₆ – Dₐ

Distance apart = 150 – 120

Distance apart = 30 Km

Therefore, car B will be 30 Km ahead of car A.

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