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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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The distance between object P1 and its image formed is determined as 36 m.

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The distance of the image formed by object P1 is calculated as follows;

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A mango falls fromthe top its tree passing a window which is 2.4m tall by taking 0.4s
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There are three points in time we need to consider.  At point 0, the mango begins to fall from the tree.  At point 1, the mango reaches the top of the window.  At point 2, the mango reaches the bottom of the window.

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y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

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Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

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0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

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0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

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3 = y₀ − 4.9 (0.412)²

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What are the conditions required for a rigid body to be in translational equilibrium?
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