Answer:
Electric potential, E = 2100 volts
Explanation:
Given that,
Electric field, E = 3000 N/C
We need to find the electric potential at a point 0.7 m above the surface, d = 0.7 m
The electric potential is given by :
V = 2100 volts
So, the electric potential at a point 0.7 m above the surface is 2100 volts. Hence, this is the required solution.
Answer:
The minimum possible coefficient of static friction between the tires and the ground is 0.64.
Explanation:
if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :
Fc = f
m×(v^2)/(R) = μ×m×g
(v^2)/(R) = g×μ
μ = (v^2)/(R×g)
= ((25)^2)/((100)×(9.8))
= 0.64
Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.
Answer:
the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
Explanation:
Given the data in the question;
we make use of the following expression;
hall Voltage VH = IB / ned
where I = 2.25 A
B = 0.685 T
d = 0.107 mm = 0.107 × 10⁻³ m
e = 1.602×10⁻¹⁹ C
VH = 2.59 mV = 2.59 × 10⁻³ volt
n is the electron density
so from the form; VH = IB / ned
VHned = IB
n = IB / VHed
so we substitute
n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )
n = 1.54125 / 4.4396226 × 10⁻²⁶
n = 3.4716 × 10²⁵ m⁻³
Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
The length of the wire is 36 m.
<u>Explanation:</u>
Given, Diameter of sphere = 6 cm
We know that, radius can be found by taking the half in the diameter value. So,
Similarly,
We know the below formulas,
When equating both the equations, we can find length of wire as below, where 
The 
value gets cancelled as common on both sides, we get
The 
value gets cancelled as common on both sides, we get
