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Levart [38]
3 years ago
13

A conducting sphere has a net charge of -4.8x10-17 C. What is the approximate number of excess electrons on the sphere?

Physics
1 answer:
Aloiza [94]3 years ago
7 0

Answer:

The number is  N  = 300

Explanation:

From the question we are told that

   The  net charge is  Q =  -4.8 *10^{-17 } \  C

Generally the charge on a electron is e = - 1.60 *10^{-19 } \ C

Generally the number of excess electrons is mathematically represented as

      N  =  \frac{Q}{e}

=>  N  =  \frac{-4.8 *10^{-17}}{-1.60 *10^{-19}}

=>  N  = 300

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andreev551 [17]

Answer:2

Explanation:

Given

Velocity of bicycle is 5 m/s towards north

radius of rim r=20 cm

mass of rim m=2 kg

Angular momentum \vec{L}=I\cdot \vec{\omega }

I=mr^2=2\times 0.2^2=0.08 kg-m^2

\omega =\frac{v}{r}=\frac{5}{0.2}=25 rad/s

L=0.08\times 25=2kg-m^2/s  

direction of Angular momentum will be towards west

7 0
3 years ago
Uncertainty in 21.0 C is<br> A. 0.1<br> B. 0.2<br> C. 0.05
san4es73 [151]

it's A.0.1

that's the answer

4 0
2 years ago
As an admirer of Thomas Young, you perform a double-slit experiment in his honor. You set your slits 1.01 mm apart and position
Snezhnost [94]

Answer:

2.316e-3 and 3.47e-3

Explanation:

Now, at that angle, we look at the bright spots on the screen.

tan θ = x / L (x is the horizontal distance from the centre of the screen, L is distance to screen)

For small angles, we can approximate that tan θ = sin θ.

nλ / d = x / L         so then       x = n λ L / d

First bright fringe, n = 1

x = (1) (641*10^-9 m) (3.65 m) / (1.01*10^-3 m) = 2.316e-3

For destructive interference (dark fringes), equation becomes:

x = (n - 0.5) λ L / d

Second dark fringe, n = 1.5

x = 1.5 (641*10^-9 m) (3.65 m) / (1.01*10^-3 m) = 3.47e-3

4 0
3 years ago
Forces that have a net force of zero the object will be still
earnstyle [38]

Answer:

No. It may be at rest or moves with a constant velocity.

Explanation:

The term "net force" refers to the vector addition of all forces acting on a defined body.

If the magnitude of the net is zero, then the acceleration of the given body would be zero. Hence, the given body may be moving with an unchanged velocity or at rest.

8 0
3 years ago
A traveling electromagnetic wave in a vacuum has an electric field amplitude of 59.3 V/m. Calculate the intensity S of this wave
bija089 [108]

Answer: S = 4.67 W/m², U = 1.29 J

Explanation:

Given

Time of flow, t = 12.3 s

Area of flow, a = 0.0225 s

Amplitude, E = 59.3 V/m

Intensity, S = ?

I = E² / cμ, where

μ = permeability of free space

c = speed of light

E = E(max) / √2

E = 59.3 / √2

E = 41.93 V/m

I = 41.93² / (2.99*10^8 * 1.26*10^-6)

I = 1758.125 / 376.74

I = 4.67 W/m²

Energy that flows through

U = Iat

U = 4.67 * 0.0225 * 12.3

U = 1.29 J

Therefore, the intensity is 4.67 W/m² and the energy is 1.29J

6 0
3 years ago
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