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attashe74 [19]
3 years ago
14

If one replaces the conducting cube with one that has positive charge carriers, in what direction does the induced electric fiel

d point
Physics
1 answer:
Grace [21]3 years ago
3 0

Answer:

There will be no change in the direction of the electric field .

Explanation:

The direction will remain the same because the sign of the charges has no effect on it.

When one replaces the conducting cube with one that has positive charge carriers there will be no change in the direction of the field as there is no defined relationship between the direction of the electric field and sign of the charge.

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skater spins over a point at a speed of 3.0 rotations per second then the momentum of inertia is 0.60 kg.M2, what is its angular
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L=11.3\ kg-m^2/s

Explanation:

Given that,

Angular speed of a skater, \omega=3\ rot/s=18.84\ rad/s

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We need to find the angular momentum of the skater. The formula for the angular momentum of the skater is given by :

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L=0.6\times 18.84\\\\L=11.3\ kg-m^2/s

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3 years ago
FREE BRAINIEST IF YOU ANSWER THIS
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The watt is a rate, similar to something like speed (miles per hour) and other time-interval related measurements.

Specifically, watt means Joules per Second. We are given that the electrical engine has 400 watts, meaning it can make 400 joules per second. If we need 300 kJ, or 3000 Joules, then we can write an equation to solve the time it would take to reach this amount of joules:

w * t = E

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(Watts times time is equal to the energy required)

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400 * t = 3000

(We need to write 3000 joules instead of 300 kilojoules, since Watts is in joules per second. It's important to make sure your units are consistent in your equations)

<u>Divide both sides by 400 to isolate t:</u>

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<u>It will take 7.5 seconds for the 400 W engine to produce 300 kJ of work.</u>

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Self-motivation is the surest way to stay focused. Nearly all the most significant tasks in life are tests for our motivation. Facing challenges and achieving success requires focus; this means we need to be responsible for our motivation. People who are motivated towards achieving their goals are focused on success. The following ideas will help you stay focused and motivated in your work and studies.

Explanation:

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2 years ago
1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x
allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }

With (\epsilon_{0}) been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    }

Explanation:

(A) Considering a uniform linear density λ_{0} on the ring, then:

dQ=\lambda dl (1)⇒Q=\lambda_{0} 2\pi R(2)⇒\lambda_{0}=\frac{Q}{2\pi R}(3)

Applying the technique of charge integration for finite charges:

V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r'  }} \, dQ(4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

r'=\sqrt{R^{2} +Z^{2}}(5)

Using the expressions (1),(4) and (5) you obtain:

V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}}  }} \, d\phi

Integrating results:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }   (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

|E| =-\nabla V (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

|E| =-\frac{dV(z)}{dz} (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    } (S_b)

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3 years ago
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