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2 years ago

Should the size of the test charge compare to the amount of charge that produces the field

Physics

1 answer:

Alla [95]2 years ago

4 0

Answer:

20000000000)0000000000000

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The tidal bulge on the side of the earth opposite the moon is due to _______ .

butalik [34]

Due to the moon's gravitational force and inertias counterbalance.

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3 years ago

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Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of

viktelen [127]

Answer:

$W=2.76\times 10^{-6}\ J$

Explanation:

Given that,

The distance between two spheres, r = 25 cm = 0.25 m

The capacitance, C = 26 pF = 26×10⁻¹² F

Charge, Q = 12 nC = 12 × 10⁻⁹ C

We need to find the work done in moving the charge. We know that, work done is given by :

$U=\dfrac{Q^2}{2C}$

Put all the values,

$U=\dfrac{(12\times 10^{-9})^2}{2\times 26\times 10^{-12}}\\\\U=2.76\times 10^{-6}\ J$

So, the work done is $2.76\times 10^{-6}\ J$ .

8 0

2 years ago

Consider the rocket in the image. Which unbalanced force accounts for the direction of the net force of the rocket ?

Wittaler [7]

Answer:

unbalenced force

Explanation:

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2 years ago

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(a) Consider a fusion generator built to create 3.00 GW of power. Determine the rate of fuel burning in grams per hour if the DT

Stels [109]

The rate of fuel burning in grams per hour if the DT reaction is used is 1.08 × $10^1^3$ J/g per hour

<h3>How is the rate of fuel burning in grams per hour calculated when the D-T reaction is used?</h3>

D + T → He + n
The D-T fusion reaction results in a Helium (He) and neutron (n)

E = 17.59 MeV

Mass = 2.014u + 3.016u

= 5.030u

Energy per Kg = (17.59× $10^6$ ×1.6× $10^-^1^9$ ) ÷ ( 5.030×1.66× $10^-^2^7$ )

= 3.37× $10^1^4$ J/Kg

= 3.0× $10^9$ J/g

Rate of fuel burning in grams per hour = 3.0× $10^9$ × 3600

= 3.6×3.0× $10^1^2$

= 1.08 × $10^1^3$ J/g per hour

To learn more about fusion reactor and energy production, refer

brainly.com/question/13399644

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4 0

1 year ago

There are two identical, positively charged conducting spheres fixed in space. The spheres are 40.4 cm apart (center to center)

Nadya [2.5K]

Answer:

$q_1=5.64\times 10^{-7}\ \text{C}$ and $q_2=2.32\times 10^{-6}\ \text{C}$

Explanation:

$F_1=0.072\ \text{N}$

$F_2=0.115\ \text{N}$

r = Distance between shells = 40.4 cm

$q_1$ and $q_2$ are the charges

$k$ = Coulomb constant = $8.99\times10^{9}\ \text{Nm}^2/\text{C}^2$

Force is given by

$F_1=\dfrac{kq_1q_2}{r^2}\\\Rightarrow q_1q_2=\dfrac{F_1r^2}{k}\\\Rightarrow q_1q_2=\dfrac{0.072\times 0.404^2}{8.99\times 10^{9}}\\\Rightarrow q_1q_2=1.307\times 10^{-12}\\\Rightarrow q_1=\dfrac{1.307\times 10^{-12}}{q_2}$

$F_2=\dfrac{kq^2}{r^2}\\\Rightarrow q=\sqrt{\dfrac{F_2r^2}{k}}\\\Rightarrow q=\sqrt{\dfrac{0.115\times 0.404^2}{8.99\times 10^{9}}}\\\Rightarrow q=1.44\times 10^{-6}\ \text{C}$

$q=\dfrac{q_1+q_2}{2}\\\Rightarrow q_1+q_2=2q\\\Rightarrow q_1+q_2=2\times1.44\times 10^{-6}\\\Rightarrow q_1+q_2=2.88\times 10^{-6}$

Substituting the above value of $q_1$ we get

$\dfrac{1.307\times 10^{-12}}{q_2}+q_2=2.88\times 10^{-6}\\\Rightarrow q_2^2-2.88\times 10^{-6}q_2+1.307\times 10^{-12}=0\\\Rightarrow \frac{-\left(-0.00000288\right)\pm \sqrt{\left(-0.00000288\right)^2-4\times \:1\times \:1.307\times 10^{-12}}}{2\times \:1}\\\Rightarrow q_2=2.32\times 10^{-6}, 5.64\times 10^{-7}$

$q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{2.32\times 10^{-6}}\\\Rightarrow q_1=5.63\times 10^{-7}$

$q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{5.64\times 10^{-7}}\\\Rightarrow q_1=2.32\times 10^{-6}$

Since we know $q_1$

$q_1=5.64\times 10^{-7}\ \text{C}$ and $q_2=2.32\times 10^{-6}\ \text{C}$ .

5 0

2 years ago