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DerKrebs [107]
3 years ago
12

In the figure below, a long circular pipe with outside radius R = 2.57 cm carries a (uniformly distributed) current i = 10.4 mA

into the page. A wire runs parallel to the pipe at a distance of 3.00R from center to center. Find the magnitude of the current in the wire in milliamperes such that the ratio of the magnitude of the net magnetic field at point P to the magnitude of the net magnetic field at the center of the pipe is 4.74, but it has the opposite direction.

Physics
2 answers:
mojhsa [17]3 years ago
8 0
It's about Ampere's Law.
docker41 [41]3 years ago
7 0
Its about Ampere's Law i think
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An ant travels 2.78 cm (West) and then turns and travels 6.25 cm (South 40 degrees East). What is the ant's total displacement?
andrey2020 [161]
From\ cosine\ theorem:\\\\
c^2=a^2+b^2-2abcos(\angle between\ a\ and\ b)\\\\
a=2,78\\b=6,25\\
\angle between\ a\ and\ b=90+50=140^{\circ}\\\\
c^2=2,78^2+6,25^2-2*2,78*6,25cos(140^{\circ})\\\\
c^2=7,7284+39,0625-34,75*(-0,77)\\\\c^2=46,7909+26,7575\\\\c^2=735484\\\\c=8,58cm\\\\Total\ displacement\ is\ equal\ to\ 8,58cm.

5 0
3 years ago
A rock is dropped from the top of a tower. When it is 40 meters above the ground velocity of 17 m/s. When its velocity is 24 m/s
eimsori [14]

Answer:

Option B is the correct answer.

Explanation:

Let us consider 40 meter above ground as origin.

Initial velocity = 17 m/s

Final velocity = 24 m/s

Acceleration = 9.81 m/s

We have equation of motion v² = u² + 2as

Substituting

         24² = 17² + 2 x 9.81 x s

           s = 14.63 m

Distance traveled by rock = 14.63 m down.

Height of rock from ground = 40 - 14.63 = 25.37 m = 25.4 m

Option B is the correct answer.

4 0
3 years ago
When you push against a wall with your fingers they bend because they experience a force what is that force
ikadub [295]

Answer:

Force = Mass * Acceleration

Explanation:

Push or Pull

3 0
3 years ago
(b) A ball is thrown from a point 1.50 m above the ground. The initial velocity is 19.5 m/s at
Nataly_w [17]

The answers are:

(i) 6.35m

(ii) 20.2 m/s


It seems like you already have the answer, but let me show you how to get it:


You have two givens:

Vi = 19.5m/s

Θ = 30°

dy = 1.50m (This is not the maximum height, just to be clear)


When working with these types of equations, you just need to know your kinematics equations. For projectiles launched at an angle, you will need to first break down the initial velocity (Vi) into its horizontal (x) and vertical (y) components.


*<em>Now remember this, if you are solving for something in the horizontal movement always use only x-components. When solving for vertical movements, always use y-components. </em>


Let's move on to breaking down the initial velocity into both y and x components.


Viy = SinΘVi = (Sin30°)(19.5m/s) = <em>9.75 m/s</em>

Vix = CosΘVi = (Cos30°)(19.5m/s) = <em>16.89 m/s</em>


Okay, so we have that down now. The next step is to decide which kinematics equation you will use. Because you have no time, you need to use the kinematic equation that is not time dependent.


(i) Maximum height above the ground


Remember that the object was thrown 1.50m above the ground. So we save that for later. First we need to solve for the maximum height above the horizontal, or the point where it was thrown.


The kinematics equation you will use is:

Vf^{2} = Vi^{2}+2ad


Where:

Vf = final velocity

Vi = inital velocity

a = 9.8m/s²

d = displacement


We will derive our displacement from this equation. And you will come up with this:

d = \dfrac{Vf^{2}-Vi^{2}}{2a}


Again, remember that we are looking for a vertical component or y-component because we are looking for HEIGHT. So we use this plugging in vertical values only.


Vf at maximum height is always 0m/s because at maximum height, objects stop. Also because gravity is a downwards force you will use -9.8m/s².

Vfy = 0 m/s a = -9.8m/s² Viy = 9.75m/s

dy = \dfrac{Vfy^{2}-Viy^{2}}{2a}

dy = \dfrac{0^{2}-(9.75m/s)^{2}}{2(-9.8m/s^{2})}

dy = \dfrac{0^{2}-(9.75m/s)^{2}}{2(-9.8m/s^{2})}

dy = \dfrac{-95.0625m^{2}/s^{2}}{-19.6m/s^{2}}

dy = 4.85m


So from the point it was thrown, it reached a height of 4.85m. Now we add that to the height it was thrown to get the MAXIMUM HEIGHT <em>ABOVE THE GROUND.</em>


4.85m + 1.50m = 6.35m


(ii) Speed before it strikes the ground. (Vf=resultant velocity)

Okay, so here we need to consider a couple of things. To get the VF we need to first figure out the final velocities of both the x and y components. We are combining them to get the resultant velocity.


Vfx = horizontal velocity = Initial horizontal velocity (Vix). This is because gravity is not acting upon the horizontal movement so it remains constant.


Vfy = ?

VF =?


We need to solve this, again, using the same formula, but this time, you need to consider we are moving downwards now. So this time, instead of Vfy being 0 m/s, Viy is now 0 m/s. This is because it started moving from rest.


Vfy^{2} = Viy^{2}+2ad

Vfy^{2} = 0m/s^{2}+2(9.8m/s^{2}(6.35m)

\sqrt{Vfy^{2}} = \sqrt{124.46m^{2}/s^{2}}

Vfy= 11.16m/s


OKAY! We are at our last step. Now to get the resultant velocity, we apply the Pythagorean theorem.


Vf^{2} = Vfx^{2} + Vfy^{2}

\sqrt{Vf^{2}} = \sqrt{(16.89m/s)^{2}+(11.16m/s)^{2}}

Vf =20.2m/s


The ball was falling at 20.2m/s before it hit the ground.

8 0
3 years ago
A car accelerates from rest to 100 m/s in 20 s. What is its acceleration?
Dmitriy789 [7]
  • initial velocity=0m/s=u
  • Final velocity=v=100m/s
  • Time=t=20s

\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \sf\longmapsto Acceleration=\dfrac{100-0}{20}

\\ \sf\longmapsto Acceleration=\dfrac{100}{20}

\\ \sf\longmapsto Acceleration=5m/s^2

7 0
3 years ago
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