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Veseljchak [2.6K]
2 years ago
8

Define fundamental and derived quantities with two example of each. ​

Physics
1 answer:
stich3 [128]2 years ago
4 0

Fundamental quantity : quantities which are independent on other physical quantity .

ex: length,mass,time, current, amount of substance, luminous intensity, thermodynamic temperature,

Derived quantity : quantities which are depend on fundamental quantities.

ex: Area, volume, density, speed, acceleration, force, velocity etc.

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2 years ago
Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y
WITCHER [35]

Question:

A point charge of -2.14uC  is located in the center of a spherical cavity of radius 6.55cm  inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm  from the center of the cavity.  

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity 3.65\times 10^5N/C

Explanation:

A point charge ,q = -2.14\times 10^{-6} C is located in the center of a spherical cavity of radius , r =6.55\times 10^{-2}  m inside an insulating spherical charged solid.  

The charge density in the solid , d = 7.35 \times 10^{-4}C/m^3.

Distance from the center of the cavity,R =9.5\times 10^{-2 }m

Volume of shell of charge= V  =(\frac{4\pi}{3})[ R^3 - r^3 ]

Charge on the shell ,Q = V \times d'

Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d

Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}

Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}

Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}

Q =1.7745 \times 10^{-6 }C

Electric field at 9.5\times 10^{-2}m due to shellE1  = \frac{k Q}{R^2}

E1 =  \frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}

E1 =1.769\times 10^6 N/C

Electric field at  9.5\times 10^{-2} due to 'q' at center E2 = \frac{kq}{R^2}

E2 =\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}

E2 =2.134\times 10^6 N/C

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

=[  2.134  - 1.769 ]\times 10^6

= 3.65\times 10^5 N/C

8 0
3 years ago
r law 1 I I hr ohm berm died I'm- an m at a r m 100 War I j V m Exam m terms of he movement of . . n molecules. how the air exer
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2 years ago
Help me plzzz<br>And thank u​
GuDViN [60]

Answer:

$ 1.1

Explanation:

From the question given above, the following data were obtained:

Cost per kWh = $ 0.1

Current (I) = 10 A

Voltage (V) = 220 V

Time (t) = 5 h

Cost of operation =?

Next, we shall determine the power the electric oven. This can be obtained as follow:

Current (I) = 10 A

Voltage = 220 V

Power (P) =?

P = IV

P = 10 × 220

P = 2200 W

Next, we shall convert 2200 W to KW. This can be obtained as follow:

1000 W = 1 KW

Therefore,

2200 W = 2200 W × 1 KW / 1000 W

2200 W = 2.2 KW

Thus, 2200 W is equivalent to 2.2 KW.

Next, we shall determine the energy consumed by the electric oven. This can be obtained as follow:

Power (P) = 2.2 KW

Time (t) = 5 h

Energy (E) =?

E = Pt

E = 2.2 × 5

E = 11 KWh

Finally, we shall determine the cost of operation. This can be obtained as follow:

1 KWh cost $ 0.1

Therefore,

11 KWh will cost = 11 × 0.1

11 KWh will cost = $ 1.1

Therefore, the cost of operating the electric oven is $ 1.1

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