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11111nata11111 [884]
3 years ago
14

a thunderclap sends a sound wave through the air and the ocean below The thunderclap sound wave has a constant frequency of 50 H

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Physics
1 answer:
Montano1993 [528]3 years ago
4 0

Answer:

child protective services

Explanation:

amonbgus

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Magnetism is the product of a moving charged particle. We can have electricity without magnetism but we can not have magnetism without electricity.An electro magnet is made so that we have a soft metal core and electricity around it. A bar magnet is a normal magnet in bar shape with permanent magnetism.
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Much of the pollution in out oceans comes from:
BabaBlast [244]

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c.) sewage and industrial waste

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Which is not an example of an external force acting on an object? (1 point)
GrogVix [38]

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A. a meteor traveling unhindered through space

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3 years ago
Two charges, each q, are separated by a distance r, and exert mutual attractive forces of F on each other. If both charges becom
jeka57 [31]

Answer:

F = ⅔ F₀

Explanation:

For this exercise we use Coulomb's law

         F = k q₁q₂ / r²

let's use the subscript "o" for the initial conditions

          F₀ = k q² / r²

now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r

       

we substitute

          F = k 4q² / 9 r²

          F = k q² r² 4/9

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3 0
3 years ago
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
Reil [10]

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

3 0
3 years ago
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