Answer:
<h2>
206.67N</h2>
Explanation:
The sum of force along both components x and y is expressed as;
The magnitude of the net force which is also known as the resultant will be expressed as 
To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.
Given the position of the object along the x-component to be x = 6t² − 4;
Similarly,
Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N
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:
Electrolytes are considered ions when placed in a solution and allow for adequate conduction of particle charges.
Explanation:
Electrolytes are substances that, when are dissolved in solution, separates into electrical positive charges (cations) and electrical negative charges (anions) which are known as ions.
These ions have an adequate capacity to conduct particle charges and, therefore electricity.
Sodium, calcium, phosphate and potassium, are examples of electrolytes.
<u>Hence, the correct answer is:</u>
Electrolytes are considered ions when placed in a solution and allow for adequate conduction of particle charges.
I hope it helps you!
Answer:
A. Zero
Explanation:
Given data,
The charge of the test charge, q = 1 C
The distance the charge moved against the filed of intensity, x = 30 cm
= 0.3 m
The electric field intensity, E = 50 N/C
The energy stored in the charge at 0.3 m is given by the formula,
V = k q/r
Where,
= 9 x 10⁹ Nm²C⁻²
The charge is moved from the potential V₁ to V₂ at 30 cm
Substituting the given values in the above equation
V₁ = 9 x 10⁹ x 30 / 0.3
= 1.5 x 10¹² J
And,
V₂ = 1.5 x 10¹² J
The energy stored in it is,
W = V₂ - V₁
= 0
Hence, the energy stored in the charge is, W = 0
To demonstrate the number of protons neutrons and electrons.
