Answer: 2.19 seconds
Explanation:
<u>Given:</u>
Initial speed, u = 0 m/s
Acceleration of the boat, a = 5 m/s^2
Distance between the boat and dock, d = 12 m
Using the third kinematics equation to solve for time:
d = u*t + (1/2)*a*t^2
12 = 0*t + (1/2)*5*t^2
t = sqrt (12*2/5)
t = 2.19 seconds
Therefore, it will take the boat approximately 2.19 seconds to reach the dock
Answer:
E(x,t) = Emaxcos(kx - ωt + φ),
B(x,t) = Bmaxcos(kx - ωt + φ).
Explanation:
E is the electric field vector, and B is the magnetic field vector of the EM wave. For electromagnetic waves the electric field E and the magnetic field B are always perpendicular to each other and perpendicular to the direction of propagation. The direction of propagation is the direction of E x B.
A change that would most improve his results would be; D) Connecting the galvanometer to the coil
Remember Coulomb's law: the magnitude of the electric force F between two stationary charges q₁ and q₂ over a distance r is
where k ≈ 8,98 × 10⁹ kg•m³/(s²•C²) is Coulomb's constant.
8.1. The diagram is simple, since only two forces are involved. The particle at Q₂ feels a force to the left due to the particle at Q₁ and a downward force due to the particle at Q₃.
8.2. First convert everything to base SI units:
0,02 µC = 0,02 × 10⁻⁶ C = 2 × 10⁻⁸ C
0,03 µC = 3 × 10⁻⁸ C
0,04 µC = 4 × 10⁻⁸ C
300 mm = 300 × 10⁻³ m = 0,3 m
600 mm = 0,6 m
Force due to Q₁ :
Force due to Q₃ :
8.3. The net force on the particle at Q₂ is the vector
Its magnitude is
and makes an angle θ with the positive horizontal axis (pointing to the right) such that
where we subtract 180° because 
terminates in the third quadrant, but the inverse tangent function can only return angles between -90° and 90°. We use the fact that tan(x) has a period of 180° to get the angle that ends in the right quadrant.
Explanation:
The magnitude of the electric field between the plates is given by
E = -ΔV/d
minus sign indicates Potential decreases in the direction of electric field
where
ΔV is the potential difference between the plates
D is the distance between the plates.
The work done when carrying an electrical charge on an equipotential surface between one position to the other is zero W= q(V-V)=0 The electric field lines of force are always perpendicular to an equipotential surface. That conductor in an equipotential surface as direction E is at right angles to an eauipotential surface The intensity of the electric field along an equipotential surface is always zero. Equipotential surfaces never collide with each other as this would mean that at that point, there are two alternative values that are not true.
