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Fed [463]
4 years ago
10

Two tiny beads are 25 cm apart with no other charges or fields present. Bead A carries 10 µC of charge and bead B carries 1 µC.

Which one of the following statements is true about the magnitudes of the electric forces on these beads?
The force on A is 10 times the force on
b. The force on A is 100 times the force on
b. The force on A is exactly equal to the force on
b. The force on B is 100 times the force on
a. The force on B is 10 times the force on
a.
Physics
1 answer:
melomori [17]4 years ago
8 0
The correct answer is (B) The force on A is exactly equal to the force on B. Two tiny beads are 25 cm apart with no other charges or fields present. Bead A carries 10 µC of charge and bead B carries 1 µC. The statement that<span> is true about the magnitudes of the electric forces on these beads is the force on A is exactly equal to B.</span>
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After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. T
Ugo [173]

Answer:

the distance is 315.3696 m

Explanation:

The computation of the distance is given below:

Given that

Sound intensity = 1.67 × 10^-6 W/m^2

And, the distance = 233 m

Now as we know that

Power = Intensity × surface area

1.67 × 10^-6 × 4π(233)^2 = 1.67 × 10^-6 ÷ 2× 4π × d^2

d^2 = 2 × (223)^2

= √2 × 223

= 315.3696 m

Hence, the distance is 315.3696 m

5 0
3 years ago
Learning Goal: To review the concept of conservative forces and to understand that electrostatic forces are, in fact, conservati
CaHeK987 [17]

Explanation:

The electrostatic forces are conservative forces!

The mainly property of the conservative fields is \vec{\nabla} \times \vec E=\vec 0

In spherical coordinates the field's expression is:

\vec E=\frac{Q}{4\pi \epsilon _0 r^2} .\^r

and the curl expression is:

\nabla\times \vec E=\frac{1}{r^2{\sin}\,\theta}\left|\begin{matrix}\hat{r} & r\,\hat{\theta} & r\,{\sin}\,\theta\,\hat{\varphi}  \\& & \\\frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \varphi}\\ & & \\E_r & rE_\theta & r{\sin}\,\theta\, E_\varphi\end{matrix}\right|=(0, 0, 0)

to find the expression for the potential function associated:

\vec E=\vec \nabla . V, \Delta V= V_b-V_a=-\int _c \vec E.d\vec l=-\int _c E\^r.dr\^r=-\int _c Edr=\int \limits^a_b \frac{Q}{4\pi \epsilon _0 r^2} dr= \frac{Q}{4\pi \epsilon _0}.(\frac{1}{r}|^b_a)= \frac{Q}{4\pi \epsilon _0}.(\frac{1}{b}-\frac{1}{a})

5 0
3 years ago
Name TWO WEAKNESSES of the model pictured below
Nikolay [14]

Answer:

Here are a few:

1) The orbital radius of these planets is ridiculously small an in no way representative of their actual radii.

2) The planets will only line up like that once every 5200 years, making this very unrepresentative of their usual relations - although this does make their order in distance from the sun.

3) The nebulae, comet, lens flare,  and other junk in the background is incorrect.

4) If this is meant as a representation of the planets, then Pluto should not be there as it is now considered a planetoid.

5) The planets are incorrectly scaled both to each other and to the sun.

7 0
3 years ago
What is the pooled variance for the following two samples?
Snowcat [4.5K]
I think it is B as 168/20
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3 years ago
Two boys with masses of 40 kg and 60 kg stand on a horizontal frictionless surface holding the ends of a light 10-m long rod. Th
kari74 [83]

When they meet the 40-kg boy will have moved a distance of 6 m.

Displacement of the 40 kg boy

The displacement of the 40 kg boy is calculated from the principle of center mass.

X(40 kg) = (60 x 10 m    +  40 x 0)/(40 kg + 60 kg)

X(40 kg) = (600)/(100) = 6 m

X(60 kg) = (60 x 0    +  40 x 10 m)/(40 kg + 60 kg)

X(60 kg) = (400)/(100) = 4 m

Thus, when they meet the 40-kg boy will have moved a distance of 6 m.

Learn more about center mass here: brainly.com/question/13499822

#SPJ1

7 0
2 years ago
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