Answer:
Charge density on the sphere = 2.2 × 10⁻⁸ C/m²
Explanation:
Given:
Radius of sphere (r) = 12 cm = 0.12 m
Distance from the electric field R = 24 cm = 0.24 m
Magnitude (E) = 640 N/C
Find:
Charge density on the sphere
Computation:
Charge on the sphere (q) = (1/K)ER² (K = 9 × 10⁹)
Charge on the sphere (q) = [1/(9 × 10⁹)](640)(0.24)²
Charge on the sphere (q) = 4 × 10⁻⁹ C
Charge density on the sphere = q / [4πr²]
Charge density on the sphere = [4 × 10⁻⁹] / [4(3.14)(0.12)²]
Charge density on the sphere = [4 × 10⁻⁹] / [0.18]
Charge density on the sphere = 2.2 × 10⁻⁸ C/m²
Answer:
the period T of whole motion should be twice the value for half at he bottom so T is 0.2sec.
w is angular frequency
formula:2π/T
now k is spring constant
F/R-->mw²
putting values:70*(2π/0.2)²
=4.9x10⁶
so we can say that SHM is not affected by the amplitude of the bounce.
Answer:
<h2><em>
15.00124mmHg</em></h2>
Explanation:
Pressure is defined as the ratio of force applied to an object to its area.
Pressure = Force/Area
Given parameters
Force = 0.242N
Area = 1.21cm²
Required parameters
Pressure = 0.242/1.21
Pressure = 0.2N/cm²
Using the conversion to convert the pressure to mmHg
1N/cm² = 75.0062mmHg
0.2N/cm² = y
y = 0.2 * 75.0062
y = 15.00124mmHg
<em>Hence the pressure in mmHg is 15.00124mmHg</em>
Answer: If one bulb goes out the other bulbs stay lit.
If there is a break in one branch of the circuit, current can still flow through the other branches.
Explanation: