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kumpel [21]
3 years ago
13

Charge is placed on the surface of a 2.7-cm radius isolated conducting sphere. the surface charge density is uniform and has the

value 6.9  10–6 c/m2. the total charge on the sphere is:
Physics
1 answer:
soldi70 [24.7K]3 years ago
5 0
First, convert the radius of the sphere from centimeter to meter.
   r = (2.7 cm)(1 m/ 100 cm) 
   r = 0.027 m

Then, calculate for the surface area of the given by the equation,
   A = 4πr²

Substituting the known values,
   A = 4(π)(0.027 m)²
   A = 9.16 x 10⁻³ m²

Then, multiply the calculated area to the charge density given to determine the total charge.
      C = (6.9 x 10⁻⁶ C/m²)(9.16 x 10⁻³ m²)
      C = 6.32 x10⁻⁸ C

<em>Answer: C = 6.32 x 10⁻⁸ C</em>

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d. Critical transportation; situation assessment; and mass care services

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4 years ago
A laser beam of unknown wavelength passes through adiffraction grating having 5510 lines/cm after striking itperpendicularly. Ta
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Answer:

Explanation:

distance between two slit  d = \frac{1 \times 10^{-2}}{5510}

d = 18.15 x 10⁻⁷ m

Let wave length of light λ

formula for  position of  first pair of bright spot

Tanθ = λ / d , λ is wave length of light and d is distance between two slit .

Tan 15.4 = \frac{\lambda}{18.15\times10^{-7}}

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4 years ago
A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.58 m/s. The pulling force is 110 N parallel to th
kari74 [83]

Given that,

Mass = 9.2 kg

Force = 110 N

Angle = 20.2°

Distance = 5.10 m

Speed = 1.58 m/s

(A). We need to calculate the work done by the gravitational force

Using formula of work done

W_{g}=mgd\sin\theta

Where, w = work

m = mass

g = acceleration due to gravity

d = distance

Put the value into the formula

W_{g}=9.2\times(-9.8)\times5.10\sin20.2

W_{g}=-158.8\ J

(B). We need to calculate the increase in internal energy of the crate-incline system owing to friction

Using formula of potential energy

\Delta U=-W

Put the value into the formula

\Delta U=-(-158.8)\ J

\Delta U=158.8\ J

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Using formula of work done

W=F\times d

Put the value into the formula

W=100\times5.10

W=510\ J

We need to calculate the work done by frictional force

Using formula of work done

W=-f\times d

W=-\mu mg\cos\theta\times d

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W=-0.4\times9.2\times9.8\cos20.2\times5.10

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We need to calculate the change in kinetic energy of the crate

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Using formula of change in kinetic energy

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v_{2}=6.35\ m/s

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(B). The increase in internal energy of the crate-incline system owing to friction is 158.8 J.

(C). The work done by 100 N force on the crate is 510 J.

(D). The change in kinetic energy of the crate is 178.7 J.

(E). The speed of the crate after being pulled 5.00m is 6.35 m/s

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