Explanation:
The quantity of charge Q in coulombs (C) that has passed through a point in a wire up to time t (measured in seconds) is given by :
We need to find the current flowing. We know that the rate of change of electric charge is called electric current. It is given by :
At t = 1 s,
Current,
So, the current at t = 1 s is 3 A.
For lowest current,
Hence, this is the required solution.
Answer:
Explanation:
Given that,
Mass m = 6.64×10^-27kg
Charge q = 3.2×10^-19C
Potential difference V =2.45×10^6V
Magnetic field B =1.6T
The force in a magnetic field is given as Force = q•(V×B)
Since V and B are perpendicular i.e 90°
Force =q•V•BSin90
F=q•V•B
So we need to find the velocity
Then, K•E is equal to work done by charge I.e K•E=U
K•E =½mV²
K•E =½ ×6.64×10^-27 V²
K•E = 3.32×10^-27 V²
U = q•V
U = 3.2×10^-19 × 2.45×10^6
U =7.84×10^-13
Then, K•E = U
3.32×10^-27V² = 7.84×10^-13
V² = 7.84×10^-13 / 3.32×10^-27
V² = 2.36×10^14
V=√2.36×10^14
V = 1.537×10^7 m/s
So, applying this to force in magnetic field
F=q•V•B
F= 3.2×10^-19 × 1.537×10^7 ×1.6
F = 7.87×10^-12 N
Work is done when spring is extended or compressed. Elastic potential energy is stored in the spring. Provided inelastic deformation has not happened, the work done is equal to the elastic potential energy stored.
Answer:
No, they will not change.
Explanation:
You need 5 blocks of the smaller object to contain the same amount of volume of the bigger object
