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2 years ago

Urgent !!!

Physics

1 answer:

arsen [322]2 years ago

8 0

The weight is not coming from the center of the mass because the force that act on it is not is equal is side.(2) section B donot have weight because the ruler bend down and section be raise up so no weight.

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How much charge is on each plate of a 3.00-Î¼F capacitor when it is connected toa 15.0-V battery? b) If this same capacitor is c

Sauron [17]

Answer:

(a) 45 micro coulomb

(b) 6 micro Coulomb

Explanation:

C = 3 micro Farad = 3 x 10^-6 Farad

V = 15 V

(a) q = C x V

where, q be the charge.

q = 3 x 10^-6 x 15 = 45 x 10^-6 C = 45 micro coulomb

(b)

V = 2 V, C = 3 micro Farad = 3 x 10^-6 Farad

q = C x V

where, q be the charge.

q = 3 x 10^-6 x 2 = 6 x 10^-6 C = 6 micro coulomb

6 0

3 years ago

What property do the following elements have in common? Li, C, and F A) They are poor conductors of electricity. B) Each element

Aloiza [94]

Answer: Option (D) is the correct answer.

Explanation:

The given elements Li, C and F are all second period elements. So, when we move from left to right across a period then there occurs increase in number of valence electrons as there occurs increase in total number of electrons.

So, it means more electrons are added to the same energy level.

Thus, we can conclude that a property of valence electrons for each element is located in the same energy level is common in the given elements.

7 0

2 years ago

Read 2 more answers

I NEED HELP PLEASE, THANKS! :)

mrs_skeptik [129]

Answer:

1. Largest force: C; smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

$F=K\dfrac{ q_{1}q_{2}}{r^{2}}$

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write

$F=\dfrac{k}{r^{2}}$

1. Net force on each particle

Let's

Call the distance between adjacent charges d.
Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

$\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}} - \dfrac{k}{(2d)^{2}} +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(b) Force on B

$\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}} - \dfrac{k}{d^{2}} + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(C) Force on C

$\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}} + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(d) Force on D

$\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}} - \dfrac{k}{(2d)^{2}} - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

$F_{A} : F_{B} : F_{C} : F_{D} = \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\$

2. Ratio of largest force to smallest

$\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}$

7 0

2 years ago

you start your bicycle ride at the top of a hill. you coast down the hill at a constant acceleration of 2.00 m/s^2. When you get

stiv31 [10]

198 meters I think, I could be wrong

6 0

2 years ago

How fast should a moving clock travel if it is to be observed by a stationary observer as running at one-half its normal rate?A)

Elan Coil [88]

Answer:

Option (D) is correct.

Explanation:

Let the speed is v.

$\Delta t = \gamma \Delta t'\\\\\Delta t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\times \frac{\Delta t}{2}\\\\\sqrt{1-\frac{v^2}{c^2}} =\frac{1}{2}\\\\1-\frac{v^2}{c^2}=\frac{1}{4}\\\\\frac{3}{4}c^2 = v^2\\\\v = 0.87 c$

Option (D) is correct.

4 0

2 years ago