Explanation:
Using Ohm's Law and a bit of substitution, we can use voltage divided by current to solve for resistance. Doing that, we'll get 6 Ohm.
Answer:
Diabetic Retinopathy is a form of diabetes that affects the eyes. It can be caused by damage to the retinas, and can cause permanent damage to the eyes, and even blindness. Initially the patient is asymptomatic and become more visibly affected in later stages. It can be treated if caught early, or in mild cases.
Explanation:
mass of the bottle in each case is M = 0.250 kg
now as per given speeds we can use the formula of kinetic energy to find it
1) when speed is 2 m/s
kinetic energy is given as
2) when speed is 3 m/s
kinetic energy is given as
3) when speed is 4 m/s
kinetic energy is given as
4) when speed is 5 m/s
kinetic energy is given as
5) when speed is 6 m/s
kinetic energy is given as
Question
A banked highway is designed for traffic moving at v 8 km/h. The radius of the curve = 330 m. 50% Part (a) Write an equation for the tangent of the highway's angle of banking. Give your equation in terms of the radius of curvature r, the intended speed of the turn v, and the acceleration due to gravity g
Part (b) what is the angle of banking of the highway? Give your answer in degrees
Answer:
a. Equation of Tangent
tan(θ) = v²/rg
b. Angle of the banking highway
θ = 0.087°
Explanation:
Given
Radius of the curve = r = 330m
Acceleration of gravity = g = 9.8m/s²
Velocity = v = 8km/h = 8 * 1000/3600
v = 2.22 m/s
a . Write an equation for the tangent of the highway's angle of banking
The Angle is calculated by
tan(θ) = v²/rg
θ = tan-1(v²/rg)
b.
Part (b) what is the angle of banking of the highway? Give your answer in degrees
θ = tan-1(v²/rg)
Substituting the values of v,g and r
θ = tan-1(2.22²/(330 * 9.8)
θ = tan-1(0.001523933209647)
θ = 0.087314873580116°
θ = 0.087°
Answer: 0.817A
Explanation:
Assuming , that one coulomb per second of negative charge alone flow through a conductor and no positive charges flow. I.e Q=It
It means a current of one A flow in the opposite direction.
This is similar to one coulomb per second of positive charge flowing through and there is no negative charge,
In addition, the one coulomb per second of positive charge flows. This is flowing in the current direction of the previous one. Then, the total current is 2 A. Since 2 coulomb of positive charges flow through one due to real positive charge and another due to the negative charge flowing in opposite direction.The charges cannot cancel each other, because even before the current flow the conductor was neutral.
According to this, the current in the given problem is
[2.7 + 2.4] x 10 ^ 18 * 1.602 x 10^ [-19] C/s
= 0.817 A
