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tankabanditka [31]
2 years ago
5

S Four identical particles, each having charge q and mass m, are released from rest at the vertices of a square of side L. How f

ast is each particle moving when their distance from the center of the square doubles?
Physics
1 answer:
SashulF [63]2 years ago
4 0

The charged particle will have velocity of 5.41 kq^2/ mL .

What is Kinetic energy ?

A particle or an item that is in motion has a sort of energy called kinetic energy. An item accumulates kinetic energy when work, which involves the transfer of energy, is done on it by exerting a net force.

Here electric potential will be :

Ui = 4k q^2 / L +2k q^2 / L root2

when we double the distance of charges the final potential is :

Uf = 2kq^2 /L + 2k Q^2/ L root2

Kinetic energy : Uf - Ui

so after putting the values and solving the equation we get :

v= 5.41 kq^2/ mL

To learn more about kinetic energy click on the link below:

brainly.com/question/12337396

#SPJ4

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A book is on the table. If the weight of the book is 25 newtons, what is the magnitude and direction of the normal force?
PilotLPTM [1.2K]

Answer: E.

Explanation: because the book is on the table, its not really moving in any direction.

7 0
3 years ago
In what way is a piano an example of a quantized system?
uranmaximum [27]

When you play a piano, your energy increases in a uniform, continuous manner and is therefore quantized.When you play a piano, you can press only on individual keys, so that your energy is restricted to certain values and is therefore quantized.

7 0
3 years ago
How high would a projectile go if it was launched from ground level with an initial speed of 26 m/s at an angle of 30 degrees ab
tigry1 [53]

Answer:

Vy = 26 m/s sin 30 = 13 m/s      vertical speed

t = Vy / a = 13 m/s / 9.80 m/s^2 = 1.33 sec     time to reach Vy = 0

H = Vy t + 1/2 g t^2

H = 13 m/s * 1.33 sec - 1.33^2 * 9.8 / 2 m = 8.62 m

4 0
11 months ago
Aliya deposited half as much money in a savings account earning 1.9% simple interest as she invested in a money market account t
yanalaym [24]

Let <em>A</em> be the amount of money that Aliya deposited in the savings account. Since <em>A</em> is half as much as money as she invested in a money market account, then the amount that she invested in the market account is <em>2A.</em>

<em />

Express the interest that Aliya earned in terms of A. Set it equal to the amount of $297.60 and then solve for <em>A</em>.

Since the savings account gives 1.9% simple interest, the total amount of interest that she will earn from the savings account is 1.9% of A, which is equal to:

\frac{1.9}{100}\times A

Since the money market account gives 3.7% simple interest, the total amount of interest that she will earn from the money market account, is 3.7% of <em>2A</em>, which is equal to:

\frac{3.7}{100}\times2A

Add both interests in terms of A and simplify the expression:

\begin{gathered} \frac{1.9}{100}\times A+\frac{3.7}{100}\times2A \\ =\frac{1.9}{100}\times A+\frac{7.4}{100}\times A \\ =(\frac{1.9}{100}+\frac{7.4}{100})\times A \\ =\frac{1.9+7.4}{100}\times A \\ =\frac{9.3}{100}\times A \end{gathered}

The expression (9.3/100)*A represents the total interest after one year. Then:

\begin{gathered} \frac{9.3}{100}\times A=297.60 \\ \Rightarrow A=\frac{100}{9.3}\times297.60 \\ \Rightarrow A=\frac{100\times297.60}{9.3} \\ \Rightarrow A=\frac{29760}{9.3} \\ \Rightarrow A=3200 \end{gathered}

Use the value of <em>A</em> to find the amount that was invested in the money market account:

2A=2\times3200=6400

Therefore, Aliya deposited 3200 in a savings account and 6400 in a money market account.

3 0
1 year ago
Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha
AlexFokin [52]

1) Electric potential inside the sphere: \frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}

where

\epsilon_0=8.85\cdot 10^{-12}F/m is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

E(r)=-\frac{dV(r)}{dr}

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

V(R)-V(r)=-\int\limits^R_r  E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)

The potential at the surface, V(R), is that of a point charge, so

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore we can find the potential inside the sphere, V(r):

V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}

On the other hand, the potential at the surface is

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore, the ratio V(center)/V(surface) is:

\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is \frac{3}{2}V(R), as found in part b) (where V(R)=\frac{Q}{4\pi \epsilon_0 R})

- Between r and R, the potential decreases as -\frac{r^2}{R^2}

- Then at r = R, the potential is V(R)

- Between r = R and r = 3R, the potential decreases as \frac{1}{R}, therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (\frac{1}{3}V(R))

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
3 years ago
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