Question

otential difference depends on current and resistance according to this function V(I,R) = IR. Your measured current and resistance have the following values and uncertainties I = 5.9 Amps, delta I space equals space 0.4 Amps and R = 42.7 Ohms and delta R space equals space 0.6 Ohms. What is the uncertainty in the , delta V ? Units are not needed in your answer.
Find the uncertainty in kinetic energy. Kinetic energy depends on mass and velocity according to this function E(m,v) = 1/2 m v2. Your measured mass and velocity have the following uncertainties delta m space equals space 0.06 kg and delta v space equals space 1.54 m/s. What is is the uncertainty in energy, delta E , if the measured mass, m = 2.27 kg and the measured velocity, v = 8.58 m/s? Units are not needed in your answer.

Find the uncertainty in a calculated electrical potential difference from the measurements of current and resistance. Electric potential difference depends on current and resistance according to this function V(I,R) = IR. Your measured current and resistance have the following values and uncertainties I = 6.6 Amps, delta I space equals space 0.6 Amps and R = 8 Ohms and delta R space equals space 0.3 Ohms. What is the uncertainty in the , delta V ? Units are not needed in your answer.

Answer 1

Answer:

Δ V =20.62

ΔE = 16.101

Δ V = 6.78

Explanation:

Case 1:

Given:

       - I = 5.9 Δ0.4  Amps

       - R = 42.7 Δ0.6  Ohms

       - V = I*R

Solution:

The procedure for finding absolute change ΔV is by taking partial derivatives of all the physical quantities.

                             V_mean = I_mean * R_mean

                            V_mean = 5.9 * 42.7 = 251.93

To find the uncertainty ΔV:

                                  ΔV / V = ΔI / I + ΔR / R

                         ΔV / 251.93 = 0.4 / 5.9 + 0.6 / 42.7

                              ΔV / 251.93 = 2062 / 25193

                                            Δ V = 20.62

Case 2:

Given:

       - m = 2.27 Δ0.06  kg

       - v = 8.58 Δ1.54  m/s

       - E = 0.5*m*v^2

Solution:

The procedure for finding absolute change ΔE is by taking partial derivatives of all the physical quantities.

                             E_mean = 0.5*m_mean *v^2_mean

                            E_mean = 0.5*2.27*8.58^2 = 83.554614

To find the uncertainty ΔE:

                                  ΔE / E = 0.5*[Δm / m + 2*Δv / v]

                         ΔE / 83.554614 = 0.5*[0.06 / 2.27 + 2*1.54 / 8.58]

                            ΔE / 83.554614= 1706 / 8853

                                            Δ E = 16.101

Case 3:

Given:

       - I = 6.6 Δ0.6  Amps

       - R = 8 Δ0.3  Ohms

       - V = I*R

Solution:

The procedure for finding absolute change ΔV is by taking partial derivatives of all the physical quantities.

                           V_mean = I_mean * R_mean

                              V_mean = 6.6 * 8 = 52.8

To find the uncertainty ΔV:

                                  ΔV / V = ΔI / I + ΔR / R

                             ΔV / 52.8 = 0.6 / 6.6 + 0.3 / 8

                                   ΔV / 52.8 = 113 / 880

                                         Δ V = 6.78