Answer:
0.0002 C.
Explanation:
Charge: This can be defined as the ratio of current to time flowing in a circuit. The S.I unit of charge is Coulombs (C)
Mathematically, charge can be expressed as
Q = CV ................................. Equation 1
Where Q = amount of charge, C = capacitance of the capacitor, V = potential difference across the plates.
Given: C = 2.0-μF = 2×10⁻⁶ F, V = 100 V.
Substitute into equation 1
Q = 2×10⁻⁶× 100
Q = 2×10⁻⁴ C
Q = 0.0002 C.
The amount of charge accumulated = 0.0002 C
Answer:
Please find the answer in the explanation
Explanation:
Given that A 1.0 g plastic bead, with a charge of -6.0 nC, is suspended between the two plates by the force of the electric field between them.
Since it is suspended, it must have been repelled by the bottom negative plate and trying to be attracted to the top plate.
We can therefore conclude that the upper plate, is positively charged
B.) The charge on the positive plate of parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates must be less than 6.0 nC
Answer:
The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J
Explanation:
The given variables are
Work done = 550 J
Volume change = V₂ - V₁ = -0.5V₁
Thus the product of pressure and volume change = work done by gas, thus
P × -0.5V₁ = 500 J
Hence -PV₁ = 1000 J
also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂
Also to compress the gas by a factor of 11 we have
P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work
Explanation:
Initial time, t₁ = 2:30 pm
Final time, t₂ = 2:30:45
We need to find the motion of students in terms of time. Final time is 45 seconds more than the initial time.
Change in time,
Hence, this is the required solution.
