Answer:
9.2 amperes
Explanation:
Ohm's law states that the voltage V across a conductor of resistance R is given by
Here, voltage V is proportional to the current I.
For voltage, unit is volts (V)
For current, unit is amperes (A)
For resistance, unit is Ohms (Ω)
Put R = 12.5 and V = 115 in V=RI
Answer:
Kinetic energy is 1425.11 J.
Explanation:
Given:
Mass of the wrench is,
Height of fall is,
Force of resistance is,
Now, the total energy at the top is equal to the potential energy of the wrench at the top since the kinetic energy at the top is 0.
Now, potential energy at the top is given as:
Now, the potential energy at the top is converted to kinetic energy at the bottom and some energy is wasted in overcoming the resistance force by air.
Potential Energy = Kinetic energy + Energy to overcome resistance.
⇒ Kinetic energy = Potential Energy - Energy to overcome resistance.
Energy to overcome resistance force is the work done by the wrench against the resistance force and is given as:
Therefore, Kinetic energy at the bottom is given as:
Hence, the kinetic energy of the wrench be when it hits the water is 1425.11 J.
Answer:
For vector u, x component = 10.558 and y component =12.808
unit vector = 0.636 i+ 0.7716 j
For vector v, x component = 23.6316 and y component = -6.464
unit vector = 0.9645 i-0.2638 j
Explanation:
Let the vector u has magnitude 16.6
u makes an angle of 50.5° from x axis
So
Vertical component
So vector u will be u = 10.558 i+12.808 j
Unit vector
Now in second case let vector v has a magnitude of 24.5
Making an angle with -15.3° from x axis
So horizontal component
Vertical component
So vector v will be 23.6316 i - 6.464 j
Unit vector of v
let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.
the mathematical formula for potential is
for positive charges the potential is positive and is negative for negative charges.
the formula for electric field is given as-
for positive charges,the line filed is away from it and for negative charges the filed is towards it.
we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.
here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.
now let the two charges of same nature.let these are positively charged.
here we can not find a point between two charges and on the line joining two charges where the potential is zero.
but at the mid point of the line joining two charges the filed is zero.