To solve this problem we will apply the concepts related to the work theorem for which it is defined as the product of Force and distance. In turn, we will use the energy conservation theorem for which the applied work must be equivalent to the total kinetic energy on the body.
The work is defined as

Here,
F = Force
d = Displacement
Replacing with our values we have that


Now by conservation of energy,



Solving for v,


Therefore the correct answer is D.
Answer:
Explanation:
The coordinate sketch for the system is shown in the attached file below. Also, in the cartesian coordinate system, since the height is less than the length and width, we did neglect the height. Thus, we eliminate the height and converted it to a two-dimension.
Answer:
1.been both -ve charged or both +be charged particles
2. 3.52mC
Explanation:
For the charge particle to cause an extension or movement of the string from its unrestrained position they would have been both -ve charged or both +be charged particles that's because like charges repel.
Now the Force sustain by the extended string is
F = Ke;
Where K is the force constant of the string, 320 N/m
e is the extension,0.033 m
F = 320 × 0.033 =10.56N
2.But according to columns law of charge;
F = kQ1 Q2
But Q1=Q2{ since the charge are of the same magnitude}.
Hence F = KQ^2
Where K is columns constant =9×10^9F/m
Hence Q=√F/K
Q= √10.56/9×10^9
=3.52×10^-3C
= 3.52mC
Τ = Iα (torque equation)
Moment of inertia of disc about axis perpendicular to the plane of disc and passing through its centre = MR²/2
∴ 12 = MR²/2 × 5.7
∴ 12 = M(0.5)²/2 × 5.7
∴ M = 12 × 2 /5.7 × 0.5² = 16.84 Kg
Answer:
2.1×10¹⁸ C
Explanation:
Using,
E = kq/r²...................... Equation 1
Where E = Electric field, q = charge, r = distance, k = coulombs constant.
make q the subject of the equation
q = Er²/k.................. Equation 2
Given: E = 180000 N/C, r = 2.8 cm = 0.028 m
Constant: k = 9×10⁹ Nm²/C².
Substitute these values into equation 2
q = 180000(9×10⁹)/0.028²
q = 2.1×10¹⁸ C
Hence the object charge is 2.1×10¹⁸ C