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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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aliya0001 [1]

Answer:

15.3 s and 332 m

Explanation:

With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon

gm = 1/6 ge

gm = 1/6 9.8 m/s² = 1.63 m/s²

We calculate the range

R = Vo² sin 2θ / g

R = 25² sin (2 30) / 1.63

R= 332 m

We will calculate the time of flight,

Y = Voy t – ½ g t2

Voy = Vo sin θ

When the ball reaches the end point has the same initial height Y=0

0 = Vo sin  t – ½ g t2

0 = 25 sin (30) t – ½ 1.63 t2

0= 12.5 t – 0.815 t2

We solve the equation

0= t ( 12.5 -0.815 t)

t=0 s

t= 15.3 s

The value of zero corresponds to the departure point and the flight time is 15.3 s

Let's calculate the reach on earth

R2 = 25² sin (2 30) / 9.8

R2 = 55.2 m

R/R2 = 332/55.2

R/R2 = 6

Therefore the ball travels a distance six times greater on the moon than on Earth

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3 years ago

What is physical change ?

ss7ja [257]

Answer:

A physical change is a change to the physical—as opposed to chemical—properties of a substance. They are usually reversible. The physical properties of a substance include such characteristics as shape (volume and size), color, texture, flexibility, density, and mass.

3 0

2 years ago

Gennadij [26K]

Answer:

3.675 m

Explanation:

$a_{x} =0$ $v_{xo}=100$ $a_{y} =-g$ $v_{yo}=0$

X-direction | Y-direction

$R=x_{o}+ v_{xo} t$ | $y=y_{o}+v_{yo}t+\frac{1}{2}a_{y}t^2$

$75=100t$ | $y=0+0+\frac{1}{2} (9.8)(0.75)$

$\frac{75}{100} =t$ | $y=3.675 m$

$0.75s=t$

Hope it helps

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3 years ago

An open pipe, 0.29 m long, vibrates in the second overtone with a frequency of 1,227 Hz. In this situation, the fundamental freq

Andru [333]

Answer:

f = 409 Hz

Explanation:

We have,

Length of the open organ pipe, l = 0.29 m

Frequency of vibration of second overtone, $f_2 = 1227 Hz$

It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :

$f_2=\dfrac{3v}{2l}$

v is speed of sound

Let f is the fundamental frequency. It is given by :

$f=\dfrac{v}{2l}$

The relation between f and f₂ can be written as :

$f_2=3f\\\\f=\dfrac{f_2}{3}\\\\f=\dfrac{1227}{3}\\\\f=409\ Hz$

So, the fundamental frequency of the pipe is 409 Hz.

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3 years ago

Your friend decides to generate electrical power by rotating a 100,000 turn coil of wire around an axis in the plane of the coil

MakcuM [25]

Answer:

a) I=35mA

b) P=1.73W

Explanation:

a) The max emf obtained in a rotating coil of N turns is given by:

$emf_{max}=NBA\omega$

where N is the number of turns in the coil, B is the magnitude of the magnetic field, A is the area and w is the angular velocity of the coil.

By calculating A and replacing in the formula (1G=10^{-4}T) we get:

$A=\pi r^2 =\pi(0.23m)^2=0.16m^2$

$emf_{max}=(100000)(0.3*10^{-4}T)(0.166m^2)(140\frac{rev}{s})=69.72V$

Finally, the peak current is given by:

$I=\frac{emf}{R}=\frac{69.72V}{1400\Omega}=49.8mA$

b)

we have that

$I_{rms}=\frac{I}{\sqrt{2}}=\frac{0.0498A}{\sqrt{2}}=0.035A$

$P_{rms}=I^2{rms}R=(0.035A)^2(1400\Omega)=1.73W$

hope this helps!!

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3 years ago