Which of the following examples show increasing potential energy?

Alexandra is attempting to drag her 32.6 kg golden retriever across the wooden floor by applying a horizontal force. What force

vichka [17]

Answer:

230.26 N

Explanation:

Since the speed is constant, acceleration is zero hence the net force will be given by the product of mass, coefficient of friction and acceleration due to gravity

F=0.72*32.6*9.81=230.26 N

7 0

2 years ago

Suppose you want to determine the resistance of a resistor that is nominally 100 . You should be able to apply 10 V across the r

Butoxors [25]

Answer:

a) For y = 102 mA, R = 98.039 ohms

For y = 97 mA, R = 103.09 ohms

b) Check explanatios for b

Explanation:

Applied voltage, V = 10 V

For the first measurement, current $y_{1} = 102 mA = 0.102 A$

According to ohm's law, V = IR

R = V/I

Here, $I = y_{1}$

$R = \frac{V}{y_{1} } \\R = \frac{10}{0.102} \\R = 98.039 ohms$

For the second measurement, current $y_{2} = 97 mA = 0.097 A$

$R = \frac{V}{y_{2} }$

$R = \frac{10}{0.097} \\R = 103 .09 ohms$

b) $y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}$

$y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]$

$y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right]$

A linear equation is of the form y = Gx

The nominal value of the resistance = 100 ohms

$x = \left[\begin{array}{ccc}100\end{array}\right]$

$\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right] = \left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] = \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5} \end{array}\right]$

3 0

2 years ago

A bowling ball traveling with constant speed hits the pins at the end of a bowling lane 16.5 m long. The bowler hears the sound

Yakvenalex [24]

Speed = distance/time
= 16.5/2.5 = 6.6 ms^-1

3 0

2 years ago

A light bulb operating at a dc voltage of 120 V has a power rating of 60 W. How much current is flowing through this bulb

densk [106]

Voltage=V=120V
Power=P=60W

Current=I

$\\ \rm\rightarrowtail I=\dfrac{P}{V}=\dfrac{60}{120}=\dfrac{1}{2}=0.5A$

8 0

2 years ago

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

maw [93]

Answer:

a) The uncertainty in calculated V, ΔV = 25.3

b) The uncertainty in calculated v, Δv = 0.41 m/s

c) The uncertainty in calculated V, ΔV = 22.2 V

Explanation:

We'll use Upper-Lower Bounds method of uncertainty to estimate the uncertainties.

a) I = 5.1 A, ΔI = 0.3 A

I = (5.1 ± 0.3) A

R = 77.5 ohms, ΔR = 0.4 ohms

R = (77.5 ± 0.4) ohms

V = IR = 5.1 × 77.5 = 395.25 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 5.1 - 0.3 = 4.8 A

Rₗ = 77.5 - 0.4 = 77.1 ohms

Vₗ = 4.8 × 77.1 = 370.08 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 5.1 + 0.3 = 5.4 A

Rᵤ = 77.5 + 0.4 = 77.9 ohms

Vᵤ = 5.4 × 77.9 = 420.66 V

The average of the differences from the mean voltage/true value is 25.3 V

V = 395.25 V, Δ = 25.3V

V = (395.25 ± 25.3) V

b) x = 2.9 m, Δx = 0.3 m

x = (2.9 ± 0.3) m

t = 4.4 s, Δt = 1.8 s

t = (4.4 ± 1.8) ohms

v = x/t = 2.9/4.4 = 0.659 m/s

The lower bound for average speed will be calculated using the lower bounds for distance and upper bounds for time.

xₗ = 2.9 - 0.3 = 2.6 m

tᵤ = 4.4 + 1.8 = 6.2 s

vₗ = 2.6/6.2 = 0.419 m/s

The upper bound for the average speed will be calculated using the upper bound for the distance and lower bound for time

xᵤ = 2.9 + 0.3 = 3.2 m

tₗ = 4.4 - 1.8 = 2.6 s

vᵤ = 3.2/2.6 = 1.231 m/s

The average of the differences from the mean average speed/true value is 0.41 m/s

v = 0.659 m/s, Δv = 0.41 m/s

v = (0.659 ± 0.41) m/s

c) ) I = 9.8 A, ΔI = 0.5 A

I = (9.8 ± 0.5) A

R = 40.5 ohms, ΔR = 0.2 ohms

R = (40.5 ± 0.2) ohms

V = IR = 9.8 × 40.5 = 396.9 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 9.8 - 0.5 = 9.3 A

Rₗ = 40.5 - 0.2 = 40.3 ohms

Vₗ = 9.3 × 40.3 = 374.79 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 9.8 + 0.5 = 10.3 A

Rᵤ = 40.5 + 0.2 = 40.7 ohms

Vᵤ = 10.3 × 40.7 = 419.21 V

The average of the differences from the mean voltage/true value is 22.2 V

V = 396.9 V, Δ = 22.2 V

V = (396.9 ± 22.2) V

7 0

2 years ago