Answer:
Minutes can provide us the continuous data as we can have the values between the two minutes, in fraction. So if we consider minutes there will be a complete continuous data for the analysis. However the number of cars, trucks, pedestrians, bicyclists, food trucks etc will be in integers as they can not be described in terms of fractions. So for continuous data option (E) is correct.
The number of minutes for a car to get from the intersection to the administration building.
Answer:
When measurements are close to true
value they are called to be accurate
Explanation:
Given Information:
Current = I = 28 A
distance between wires = r = 2.0 m
Required Information:
Magnetic field = B = ?
Answer:
B = 12x10⁻⁶ T
Step-by-step explanation:
Biot-Savart Law is given by
B = μ₀I/2πr
Where μ₀ is the permeability of free space, I is the current flowing through the wire and B is the magnitude of the magnetic field produced.
We are asked to find the magnetic field midway between the wires so r/2 = 1
B = 4πx10⁻⁷*28/2π*1
B = 6x10⁻⁶ T
since the same amount of current flows in both wires therefore, equal amount of magnetic field will be produced in both wires
B = 2*6x10⁻⁶ T
B = 12x10⁻⁶ T
Therefore, the net magnetic field midway between the two wires is 12x10⁻⁶ T.
Answer:
The variation rate is 5.42 10⁻⁵ cm²/ºC
Explanation:
When we have a thermal expansion problem we must have the relationship of the change in length as a function of the temperature, which are given in this problem, so we can write the expression for the area of a rectangle
a = L W
They ask us to find the rate of variation of this area depending on the temperature, so we can derive this expression with respect to the temperature
da / dT = d(LW) / dt
We use the derivative of a product since the two magnitudes change
da / dT = W dL/dT + L dW/dT
The values they give us are
= 1.9 10⁻⁵ cm/ºC
= 8.5 10⁻⁶ cm/ºC
W = 1.6 cm
L= 2.8 cm
Substituting the values and calculating
= 1.6 1.9 10⁻⁵ + 2.8 8.5 10⁻⁶
= 3.04 10⁻⁵ + 2.38 10⁻⁵
= 5.42 10⁻⁵ cm²/ºC
The variation rate is 5.42 10⁻⁵ cm²/ºC
