Answer:
Explanation:
Given:
- Total initial charge on bead A=26 nC
- The distance between them=5 cm
- Magnitude o the force between them
Using coulombs law the force between two charged particle is 
where r is the radial distance between them
According to question we have
Hence the charge on two metal beads is calculated.
Answer:
Option C is the correct answer.
Explanation:
By Charles's law we have
V ∝ T
That is
Here given that
V₁ = 0.20 cubic meter
T₁ = 333 K
T₂ = 533 K
Substituting
New volume of the gas = 0.3198 m³
Option C is the correct answer.
For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s
Answer:
Net charge contained in the cubeq= 3.536×10^-6C
Explanation:
Formular for total flux in a cube is given as:
Total flux= E300Acos(180) + E200Acos(0)
Where A is crossectional area
Total flux= A(E200-E300)
Total flux= q/Eo
q= Eo×total flux
q=(8.84×10^-12)×(100)^2×(100-60)
q= 3.536×10^-6C
