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irga5000 [103]
3 years ago
9

Find the uncertainty in a calculated electrical potential difference from the measurements of current and resistance. Electric p

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.
Physics
1 answer:
anygoal [31]3 years ago
3 0

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

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a substance dissolves.  


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Fill in the blanks for the following:
storchak [24]

Answer:

<em>a. 4.21 moles</em>

<em>b. 478.6 m/s</em>

<em>c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>

Explanation:

Volume of container = 100.0 L

Temperature = 293 K

pressure = 1 atm = 1.01325 bar

number of moles n = ?

using the gas equation PV = nRT

n = PV/RT

R = 0.08206 L-atm-mol^{-1}K^{-1}

Therefore,

n = (1.01325 x 100)/(0.08206 x 293)

n = 101.325/24.04 = <em>4.21 moles</em>

The equation for root mean square velocity is

Vrms = \sqrt{\frac{3RT}{M} }

R = 8.314 J/mol-K

where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol

Vrms = \sqrt{\frac{3*8.314*293}{0.0319} }= <em>478.6 m/s</em>

<em>For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship</em>

\frac{Voxy}{Vnit} = \sqrt{\frac{Mnit}{Moxy} }

where

Voxy = root mean square velocity of oxygen = 478.6 m/s

Vnit = root mean square velocity of nitrogen = ?

Moxy = Molar mass of oxygen = 31.9 g/mol

Mnit = Molar mass of nitrogen = 14.00 g/mol

\frac{478.6}{Vnit} = \sqrt{\frac{14.0}{31.9} }

\frac{478.6}{Vnit} = 0.66

Vnit = 0.66 x 478.6 = <em>315.876 m/s</em>

<em>the root mean square velocity of the oxygen gas is </em>

<em>478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>

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