Answer:
3.135 kN/C
Explanation:
The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}
Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m
So, E = qz/{4πε₀[√(z² + R²)]³}
E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}
E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}
E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}
E = 6.839 × 10³ Cm²/[1.297 m]³F
E = 6.839 × 10³ Cm²/2.182 m³F
E = 3.135 × 10³ V/m
E = 3.135 × 10³ N/C
E = 3.135 kN/C
Answer:
This property could be used to create technologically-advanced tools or machines that could easily locate the mineral deposits.
Explanation:
Mineral deposits are hard to find, unless you have the skill or the proper tools in locating them. This is the reason why many people are mining in order to explore the different areas where they could find these deposits.
If one would consider the property of minerals, such as being good conductors of heat and electricity,<u> then they could create a tool or machine that would aid in their exploration.</u> Inventors could probably come up with a sensitive detector which signals when it reaches an area of high heat and electric conductivity. Since most minerals such as <em>gold, silver, copper, galena, bornite </em>and the like have this property, then miners will have a lesser amount of time looking for them.
If this technology will be implemented, though, regulation policy must be strictly implemented because it might lead to<em> over-mining</em> thus leading to the depletion of mineral deposits.
Assuming an ideal gas, the speed of sound depends on temperature
only. Air is almost an ideal gas.
Assuming the temperature of 25°C in a "standard atmosphere", the
density of air is 1.1644 kg/m3, and the speed of sound is 346.13 m/s.
The velocity can't be specified, since the question gives no information
regarding the direction of the sound.
Answer:
C
Explanation:
- Let acceleration due to gravity @ massive planet be a = 30 m/s^2
- Let acceleration due to gravity @ earth be g = 30 m/s^2
Solution:
- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:
t = v / a
t = v / 30
- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:
t = v / g
t = v / 9.81
- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C
Answer:
200 mL
Explanation:
Given that,
Initial volume, V₁ = 300 mL
Initial pressure, P₁ = 0.5 kPa
Final pressure, P₂ = 0.75 kPa
We need to find the final volume of the sample if pressure is increased at constant temperature. It is based on Boyle's law. Its mathematical form is given by :
V₂ is the final volume
So, the final volume of the sample is 200 mL.
