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Gre4nikov [31]
3 years ago
14

uniformly charged conducting sphere of 1.2 m diameter has surface charge density 8.1 mC/m2. Find (a) the net charge on the spher

e and (b) the total electric flux leaving the surface.
Physics
1 answer:
Stels [109]3 years ago
5 0

Answer:

(a) = 3.7 × 10⁻⁵

(b) = 4.1 × 10⁻⁶ N.M²/C

Explanation:

(a) Diameter of the sphere, d = 1.2 m

Radius of the sphere, r = 0.6 m

Surface charge density, = 8.1 mC/m2 = 8.1 × 10⁻⁶ C/m²

Total charge on the surface of the sphere,

Q = Charge density × Surface area

    = 4πr²σ

   = 4 (3.14) (0.12²) (8.1 × 10⁻⁶)

   = 3.66 × 10⁻⁵C

   ≅ 3.7 × 10⁻⁵C

Therefore, the net charge on the sphere is  3.7 × 10⁻⁵C

(b)

Total electric flux (∅)

=Q / ε₀

ε₀ =  8.854 × 10⁻¹² N⁻¹C² m⁻²

Q =   3.66 × 10⁻⁵C

 =  3.66 × 10⁻⁵ / 8.854 × 10⁻¹²

= 4.1 × 10⁻⁶ N.M²/C

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When a cold alcohol-in-glass thermometer is first placed in a hot tub of water, the alcohol initially descends a bit and then ri
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3 years ago
At what rate would heat be lost through the window if you covered it with a 0.750 mm-thick layer of paper (thermal conductivity
igor_vitrenko [27]

The given question is incomplete. The complete question is as follows.

A picture window has dimensions of 1.40 m, 2.50 m and is made of glass 6.00 mm thick. On a winter day, the outside temperature is -15^{o}C, while the inside temperature is a comfortable 21.0^{o}C.

Part A

At what rate is heat being lost through the window by conduction?

Express your answer using three significant figures.

Part B

At what rate would heat be lost through the window if you covered it with a 0.750 mm-thick layer of paper (thermal conductivity 0.0500 W/m?K)?

Express your answer using three significant figures.

Explanation:

Formula for rate of heat transferred through single thick plane of glass is as follows.

       (\frac{Q}{\Delta t)}_{single} = \frac{A(T_{h} - T_{c})}{\frac{L_{ghss}}{K_{ghss}}}

                               = \frac{1.4 \times 2.5 m^{2} \times 36^{o}C}{\frac{6.0 mm}{0.8 W/m K}}

                           = 16.8 \times 10^{3} W

When window is covered by paper then the rate of heat transfer is as follows.

(\frac{Q}{\Delta t)}_{single} = \frac{A(T_{h} - T_{c})}{\frac{L_{ghss}}{K_{ghss}} + \frac{L_{paper}}{K_{paper}}}

                 = \frac{(1.4 \times 2.5 m^{2} \times 36^{o}C}{\frac{6.0 mm}{0.8 W/m K} + \frac{0.75 mm}{0.05 W/m K}}

                 = 5.6 \times 10^{3} W

Thus, we can conclude that heat lost is 5.6 \times 10^{3} W.

3 0
4 years ago
A skater increases her speed from 5.0 m/s to 10 m/s in 5 seconds. What is the acceleration of the skater?
forsale [732]

Answer:

1 m/s^2

Explanation:

The acceleration of an object is the rate of change of velocity of the object; it is given by:

a=\frac{v-u}{t}

where

u is the initial velocity of the body

v is its final velocity

t is the time elapsed

In this problem, we have:

u = 5.0 m/s is the initial velocity of the skater

v = 10.0 m/s is the final velocity

t = 5 s is the time elapsed

Therefore, the acceleration of the skater is:

a=\frac{10-5}{5}=1 m/s^2

8 0
4 years ago
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