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eimsori [14]
3 years ago
11

A long, straight metal rod has a radius of 5.70 cm and a charge per unit length of 28.2 nC/m. Find the electric field at the fol

lowing distances from the axis of the rod, where distances are measured perpendicular to the rod's axis. (a) 3.10 cm magnitude N/C direction (b) 18.0 cm magnitude N/C direction(c) 180 cm magnitude N/C direction
Physics
1 answer:
tekilochka [14]3 years ago
5 0

Answer:

0

2817.43213 N/C

281.74321 N/C

Explanation:

The metal rod is conducting electricity. This means that the charges are being carried on the surface. Inside the rod there is no electric field. So, the electric field intensity is zero at a distance 3.1 cm.

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

r = Distance

\lambda = Charge per unit length = 28.2 nC/m

Electric field is given by

E=\dfrac{\lambda}{2\pi \epsilon_0r}\\\Rightarrow E=\dfrac{28.2\times 10^{-9}}{2\pi \times 8.85\times 10^{-12}\times 0.18}\\\Rightarrow E=2817.43213\ N/C

The electric field intensity is 2817.43213 N/C

E=\dfrac{\lambda}{2\pi \epsilon_0r}\\\Rightarrow E=\dfrac{28.2\times 10^{-9}}{2\pi \times 8.85\times 10^{-12}\times 1.8}\\\Rightarrow E=281.74321\ N/C

The electric field intensity is 281.74321 N/C

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Can you please include the statement or the model?
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2 years ago
A basketball player can leap upward 0.43 m. how long does he remain in the air? use an acceleration due to gravity of 9.80 m/s2
MArishka [77]
From the equations of linear motion,
v² = u² + 2as where v is the final velocity, u is the initial velocity and a is the gravitational acceleration, and s is the displacement,
Thus, v² = u² -2gs, but v=0
hence, u² = 2gs
                = 2×9.81×0.43
                = 8.4366
            u = √8.4366
               =2.905 m/s
Hence the initial velocity is 2.905 m/s
 Then using the equation v= u +gt .
Therefore, v = u -gt. (-g because the player is jumping against the gravity)
but, v = 0
Thus, u= gt
Hence, t = u/g
              = 2.905/9.81
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3 0
3 years ago
Lila is a track and field athlete. She must complete four laps around a circular track. The track itself is a 400 meter track an
Aleonysh [2.5K]

Answer:

Her speed is 1.1 m/s, and her velocity is 0 m/s

Explanation:

Speed = Distance covered/Time

Given

Distance = 400m

Time = 6minutes = 6*60 = 360 secs

Substitute the given parameter into the formula;

Speed = 400/360

Speed = 1.1m/s

Since the track is a circular track, the displacement will be zero. She is only moving in a circular path (no direction)

Velocity = Displacement/Time

Velocity = 0/3600

Velocity = 0m/s

Hence her speed is 1.1 m/s, and her velocity is 0 m/s

6 0
3 years ago
4. As Juan is going to take a shower, the soap falls out of the soap dish on to the
LiRa [457]

The coefficient of friction between the soap and the floor is 0.081

If Juan steps on the soap with a force of 493 N, this is her weight, W. This weight also equals the normal reaction on the floor, N.

We know that frictional force F = μN where μ = coefficient of friction between soap and floor.

So, μ = F/N

Since F = 40 N and N = W = 493 N,

μ = F/N

μ = 40 N/493 N

μ = 0.081

So, the coefficient of friction between the soap and the floor is 0.081

Learn more about coefficient of friction here:

brainly.com/question/13923375

5 0
3 years ago
An object is moving along a straight line, and the uncertainty in its position is 1.90 m.
just olya [345]

Answer:

2.78\times 10^{-35}\ \text{kg m/s}

6.178\times 10^{-34}\ \text{m/s}

0.31\times 10^{-4}\ \text{m/s}

Explanation:

\Delta x = Uncertainty in position = 1.9 m

\Delta p = Uncertainty in momentum

h = Planck's constant = 6.626\times 10^{-34}\ \text{Js}

m = Mass of object

From Heisenberg's uncertainty principle we know

\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}

The minimum uncertainty in the momentum of the object is 2.78\times 10^{-35}\ \text{kg m/s}

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m=0.045\ \text{kg}

Uncertainty in velocity is given by

\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}

The minimum uncertainty in the object's velocity is 6.178\times 10^{-34}\ \text{m/s}

Electron

m=9.11\times 10^{-31}\ \text{kg}

\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}

The minimum uncertainty in the object's velocity is 0.31\times 10^{-4}\ \text{m/s}.

6 0
2 years ago
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