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Sindrei [870]
2 years ago
12

100 Pascal is equal to: A) 100 Newton B) 1000N/m2 C) 100N/m2 D) 10,000N/m2 E) None of these

Physics
2 answers:
Dmitrij [34]2 years ago
8 0
E i thiiinnnlkkkkkkkk
irina1246 [14]2 years ago
4 0

Answer: Choice C)  100 N/m^2

=======================================================

Work Shown:

1 pascal = 1 N/m^2

100 pascals = 100 N/m^2

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A student calculates the density of a copper cube to be 4.15 g/cm . If the accepted value is 8.64 g/cm the percentage error in h
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 The percentage error in his experimental value is -51.97%.

<h3>What is percentage error?</h3>

This is the ratio of the error to the actual measurement, expressed in percentage.

To calculate the percentage error of the student, we use the formula below.

Formula:

  • Error(%) = (calculated value-accepted value)100/(accepted............. Equation 1

From the question,

Given:

  • Calculated value = 4.15 g/cm
  • accepted value = 8.64 g/cm

Substitute these values into equation 1

  • Error(%) = (4.15-8.64)100/8.64
  • Error(%) = -4.49(100)/8.64
  • Error(%) = -449/8.64
  • Error(%) = -51.97 %

 

Hence, The percentage error in his experimental value is -51.97%.

Learn more percentage error here: brainly.com/question/5493941

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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

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