Answer:81.235N
Explanation:
Work=88J
theta=10°
distance=1.1 meters
work=force x cos(theta) x distance
88=force x cos10 x 1.1 cos10=0.9848
88=force x 0.9848 x 1.1
88=force x 1.08328
Divide both sides by 1.08328
88/1.08328=(force x 1.08328)/1.08328
81.235=force
Force=81.235
The correct answer for the question that is being presented above is this one: "B.pushing against a car without moving it." According to the scientific definition, pushing against a car without moving it is not an example of work. Lifting a book off a desk and <span>pulling socks out of the drye are samples of work.</span>
Answer:
A) μ = A.m²
B) z = 0.46m
Explanation:
A) Magnetic dipole moment of a coil is given by; μ = NIA
Where;
N is number of turns of coil
I is current in wire
A is area
We are given
N = 300 turns; I = 4A ; d =5cm = 0.05m
Area = πd²/4 = π(0.05)²/4 = 0.001963
So,
μ = 300 x 4 x 0.001963 = 2.36 A.m².
B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;
B = (μ_o•μ)/(2π•z³)
Let's make z the subject ;
z = [(μ_o•μ)/(2π•B)] ^(⅓)
Where u_o is vacuum permiability with a value of 4π x 10^(-7) H
Also, B = 5 mT = 5 x 10^(-6) T
Thus,
z = [ (4π x 10^(-7)•2.36)/(2π•5 x 10^(-6))]^(⅓)
Solving this gives; z = 0.46m =
Answer:
b. Constant magnitude, but varying direction, perpendicular to the equipotential.
Explanation:
As we know that the relation between electric field and electric potential is given as
here if we say that potential is constant because electric field sensor is moving along equi-potential line.
Then we will say
V = constant
so we have
so electric field will remain constant always in magnitude and always remains perpendicular to the surface
so we have
b. Constant magnitude, but varying direction, perpendicular to the equipotential.
