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Tcecarenko [31]
3 years ago
13

Problem 4: a long wire carries current towards east. a positive charge moves westward and just north from the wire. what is the

direction of the force experienced by this charge?
Physics
1 answer:
Alex787 [66]3 years ago
4 0
The direction of the force experienced by the positive charge is upward.

We can use the right-hand rule to understand the direction of the Lorentz force acting on the charge: let's put the thumb in the same direction of the current in the wire (eastward), while the other fingers "wrap themselves" around the wire. These other fingers give the direction of the Lorentz force in every point of the space around the wire. Since the charge is located north of the wire, in that point the fingers are directed upward, so the positive charge experiences a force directed upward.
(if it was a negative charge, we should have taken the opposite direction)
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Modern observations have shown that the geometry of the universe is ____.
lesantik [10]

Modern observations have shown that the geometry of the universe is flat and the universe mist be infinite.

<h3>What is geometry?</h3>

Geometry refer to a branch of mathematics that encapsulates measurements of lines, shapes, dimensions, points and shapes.

Therefore, Modern observations have shown that the geometry of the universe is flat and the universe mist be infinite.

Learn more about geometry below.

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8 0
1 year ago
A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along
storchak [24]

Answer:

v = \frac{kQ}{a}  

Explanation:

We define the linear density of charge as:

\lambda = \frac{Q}{L}

     Where L is the rod's length, in this case the semicircle's length L = πr

The potential created at the center by an differential element of charge is:

dv = \frac{kdq}{r}

          where k is the coulomb's constant

                     r is the distance from dq to center of the circle

Thus.

v = \int_{}^{}\frac{kdq}{a}  

v = \frac{k}{a}\int_{}^{}dq

v = \frac{kQ}{a}     Potential at the center of the semicircle

4 0
3 years ago
Which of these is a part of the biosphere? a. wind b. lakes c. bacteria d. glaciers
Aleks [24]
B. lakes is a part of the biosphere
Biosphere is defined as "<span>the regions of the surface, atmosphere, and hydrosphere of the earth (or analogous parts of other planets) occupied by living organisms." and a lake is the only thing on the list that could be occupied by a living thing.</span>
4 0
3 years ago
Read 2 more answers
Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init
Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ=(m+\frac{1}{2})\frac{lamda}{n}  ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= \frac{3*lamda}{2}

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= \sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }

δ= \sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}

δ= \sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }

δ= \sqrt{4900+43^2}-\sqrt{4900+23^2}

δ= \sqrt{4900+1849}-\sqrt{4900+529}

δ= \sqrt{6749}-\sqrt{5429}

δ=  82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= \frac{3*lamda}{2}

δ= 8.47

8.47 = \frac{3*lamda}{2}

λ = \frac{8.47*2}{3}

λ = 5.65m

3 0
3 years ago
If the frog lands with a velocity equal to its average velocity and comes to a full stop 0.25s later, what is the frog’s average
Nina [5.8K]

Answer:

Average accelation = -4V

Explanation:

a=\frac{V-V0}{t}

V=0 m/s (because the frog stopped)

V0 = V (average velocity)

t= 0,25 s

So;

a=\frac{V-V0}{t}=\frac{0-V}{0.25}=-4V

4 0
3 years ago
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