A transverse and longitudinal wave combine to form what kind of wave?

Two tiny conducting spheres are identical and carry charges of -19.8μC and +40.7μC. They are separated by a distance of 3.59 cm.

romanna [79]

Answer:

(a): $\rm -5.627\times 10^3\ N.$

(b): $\rm 7.626\times 10^2\ N.$

Explanation:

Given:

Charge on one sphere, $\rm q_1 = -19.8\ \mu C = -19.8\times 10^{-6}\ C.$

Charge on second sphere, $\rm q_2 = +40.7\ \mu C = +40.7\times 10^{-6}\ C.$
Separation between the spheres, $\rm r=3.59\ cm = 3.59\times 10^{-2}\ m.$

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two static point charges is given by

$\rm F=k\cdot\dfrac{q_1q_2}{r^2}$

where,

k is called the Coulomb's constant, whose value is $\rm 9\times 10^9\ Nm^2/C^2.$

From Newton's third law of motion, both the spheres experience same force.

Therefore, the magnitude of the force that each sphere experiences is given by

$\rm F=k\cdot\dfrac{q_1q_2}{r^2}\\=9\times 10^9\times \dfrac{(-19.8\times 10^{-6})\times (+40.7\times 10^{-6})}{(3.59\times 10^{-2})^2}\\=-5.627\times 10^3\ N.$

The negative sign shows that the force is attractive in nature.

Part (b):

The spheres are identical in size. When the spheres are brought in contact with each other then the charge on both the spheres redistributes in such a way that the net charge on both the spheres distributed equally on both.

Total charge on both the spheres, $\rm Q=q_1+q_2=-19.8\ \mu C+40.7\ \mu C = 20.9\ \mu C.$

The new charges on both the spheres are equal and given by

$\rm q_1'=q_2'=\dfrac Q2 = \dfrac{20.9}{2}\ \mu C=10.45\ \mu C = 10.45\times 10^{-6}\ C.$

The magnitude of the force that each sphere now experiences is given by

$\rm F'=k\cdot \dfrac{q_1'q_2'}{r^2}'\\=9\times 10^9\times \dfrac{10.45\times 10^{-6}\times 10.45\times 10^{-6}}{(3.59\times 10^{-2})^2}\\=7.626\times 10^2\ N.$

7 0

3 years ago

if a ball on a string swung in a circles and string suddenly breaks.the ball will move into what direction?

Akimi4 [234]

Answer: b

Explanation: did the quiz got it right

3 0

3 years ago

Select the correct answer. Which chemical reaction absorbs energy? A. photosynthesis B. explosion C. current produced by a batte

zavuch27 [327]

Answer:

Explanation:

Photosynthesis

4 0

3 years ago

Read 2 more answers

A 70.0-kg skier is sliding at 4 m/s when they slide down a 2m high hill. At the bottom of the hill they run into a large 2800 N/

Katarina [22]

Answer:

$\Delta x=245\ mm$

Explanation:

Given:

mass of skier, $m=70\ kg$

initial velocity of skier, $u=4\ m.s^{-1}$

height of the hill, $h=2\ m$

spring constant, $k=2800\ N.m^{-1}$

final velocity of skier before coming in contact of spring:

Using eq. of motion:

$v^2=u^2+gh$

$v^2=4^2+9.8\times 2$

$v=5.9666\ m.s^{-1}$

Now the time taken by the skier to reach down:

$v=u+gt$

$5.9666=4+9.8\ t$

$t=0.2007\ s$

Now we calculate force using Newton's second law:

$F=\frac{dp}{dt}$

$F=\frac{m(v-u)}{t}$

$F=\frac{70\times(5.9666-4)}{0.2007}$

$F\approx686\ N$

∴Compression in spring before the skier momentarily comes to rest:

$\Delta x=\frac{F}{k}$

$\Delta x=\frac{686}{2800}$

$\Delta x=0.245\ m$

$\Delta x=245\ mm$

4 0

3 years ago

1) Consider a source particle of charge qS=9 C located at (−9,5) [distances in meters]. Find the electric field vector at the ta

xeze [42]

Answer:

1) E = 2.25 i^+ 0.809j^) 10⁹ N / C , 2) E = 2.39 10⁹ N / C , 3) θ = 19.8º , 4) F = 19.12 10⁹ N , 5) E = (1.32 i^+ 3.56 j^) 109 N/C

Explanation:

1) The equation for the electric field is

E = k q / r²

Where K is the Coulomb constant that is worth 8.99 10⁹ N m² /C², q is the load and r is the distance of the load to the test point

Since the electric field is a vector magnitude, we can find its component

X axis

Ex = k q / x²

where the distance on the axis is

x = √ (X₂-x₁)²

x = √ (-15 + 9)² = 6 m

Eₓ = 8.99 10⁹ 9/6²

Eₓ = 2.25 10⁹ N /C

Y Axis

y = √ (y₂-y₁)² = √ (15-5)² = 10 m

$E_{y}$ = 8.99 10⁹ 9/10²

$E_{y}$ = 0.809 10⁹ N / C

E = Eₓ i^+ $E_{y}$ j^

E = 2.25 i^+ 0.809j^) 10⁹ N / C

2) the magnitude can be found using the Pythagorean triangle

E = √ (Eₓ² + $E_{y}$ ²)

E = √ (2.25² + 0.809²) 10⁹

E = 2.39 10⁹ N / C

3) to find the angle let's use trigonometry

tan θ = $E_{y}$ / Eₓ

θ = tan⁻¹ $E_{y}$ / Eₓ

θ = tan⁻¹ (0.809 / 2.25)

θ = 19.8º

Regarding the positive side of the x axis

4) a charge q2 = 8C is placed, let's calculate the force

F = q E

F = 8 2.39 10⁹

F = 19.12 10⁹ N

5) The total electric field at the origin, let's look for its components

q₁ = 9C

r₁ = -9 i ^ + 5 j ^

q₂ = 8 C

r₂ = -15 i ^ + 15 j ^

X axis

Eₓ = E₁ₓ + E₂ₓ

Eₓ = k q₁ / Dx₁² + k q₂ / Dx₂²

Eₓ = 8.99 10⁹ (9 / (9-0)² + 8 / (15-0)²)

Eₓ = 1.32 109 N / A

Y Axis

$E_{y}$ = $E_{1y}$ + $E_{2y}$

$E_{y}$ = k q₁ / Δy₁² + k q₂ / Δy₂²

$E_{y}$ = 8.99 109 (9/5² + 8/15²)

$E_{y}$ = 3.56 109 N / A

E = (1.32 i^+ 3.56 j^) 109 N/C

7 0

4 years ago