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alexdok [17]
2 years ago
15

A spherical capacitor is formed from two concentric sphericalconducting shells separated by vacuum. The inner sphere has radius1

0.0 centimeters, and the separation between the spheres is 1.50centimeters. The magnitude of the charge on each sphere is 3.30nanocoulombs.1)What is the magnitude of the potential difference DeltaV between the two spheres?2)What is the electric-field energy stored in the capacitor?
Physics
1 answer:
Tema [17]2 years ago
7 0

Answer:

Δ V = 38.74 V, W= 6.39 × 10 ⁻⁸ J

Explanation:

Given: Let inner Sphere Radius r1 = 10.0 cm = 0.1 m

separation d = 1.50 cm = 0.015 m

so the radius of outer sphere will be 0.1 + 0.015 = 0.115 m =r2

charge Q = 3.30 nC= 3.30 × 10 ⁻⁹ C

K= 9.0 x 10⁹ N • m² / C²

To Find: Δ V= ? and W=?

Solution:

1) we know that   Δ V = K Q / (1/r1 - 1/r2)

Δ V = 9.0 x 10⁹ N • m² / C² × 3.30 × 10 ⁻⁹ C / (1/0.1 - 1/0.115)

Δ V = 38.74 V

2) W = VQ/2    

W=   1980 V  × 3.30 × 10 ⁻⁹ C /2

W= 6.39 × 10 ⁻⁸ J

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Answer:

8616.7468 \ kg/m^3

Explanation:

Pressure is measured is p=\rho gh here p is pressure \rho is density and h is height

We have given pressure p=9.891\times 10^4\ Pa acceleration due to gravity g=9.9870\ m/sec^2 height =1.163 m

\rho =\frac{p}{gh}=\frac{9.891\times 10^4}{9.870\times 1.163}=8616.7468 \ kg/m^3

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2 years ago
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Magnus has reached the finals of a strength competition. In the first round, he has to pull a city bus as far as he can. One end
iragen [17]

Answer:

The workdone is  W_d =-4400J

Explanation:

The free body diagram is shown on the first uploaded image

From the question we are given that

            The force is on the force gauge  F = 2750 N

             The distance that Magnus pulled the bus  d = 1.60m

Generally  the workdone by the tension force on Magnus is

                  Workdone = Force * displacement \ in \ the \ direction \ of \ force

                     W_d = F * (-d)

This negative sign show that is tension force  is in the opposite direction to Magnus movement (i.e the movement of the bus )

Substituting value we have

                   Workdone  =  - 2750 * 1.60

                                     =-4400 J

7 0
2 years ago
Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci
Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W

\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})

h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}

2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99

6 0
2 years ago
A car travels 500m in 50s, then 1,500m in 75s. Calculate its averages speed for the whole journey
SIZIF [17.4K]

Answer:

15m/s

Explanation:

500 ÷ 50 = 10m/s

1500 ÷ 75 = 20m/s

10 + 20 = 30

30 ÷ 2 = 15m/s

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Answer:

2666 kg

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Explanation:

m = Mass of boat

a = Acceleration of boat

From Newton's second law

Force

F=ma\\\Rightarrow F=215\times 1.55\\\Rightarrow F=333.25\ N

Force on the first boat is 333.25 N

F=ma\\\Rightarrow m=\frac{F}{a}\\\Rightarrow m=\frac{333.25}{0.125}\\\Rightarrow m=2666\ kg

Hence, mass of the second boat is 2666 kg

Combined mass = 2666+215 = 2881 kg

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The acceleration on the combined mass is 0.11567 m/s²

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