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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?

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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Kinetic energy lost in collision is 10 J.

<u>Explanation:</u>

Given,

Mass, $m_{1}$ = 4 kg

Speed, $v_{1}$ = 5 m/s

$m_{2}$ = 1 kg

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Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

$\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2$

By plugging in the values we get,

$KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\$

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

$KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)$

$KE = 40J + KE(lost)$

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

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Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat

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Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

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Differentiate with respect to t

$\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}$

$\Rightarrow 10 =4\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}$

The surface of area of the balloon is(S) = $4\pi r^2$

$S=4\pi r^2$

Differentiate with respect to t

$\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}$

Putting the value of $\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}$

$\Rightarrow \frac{dS}{dt} =\frac{20}{ r}$

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Answer:

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Now,buyantant force

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