All categories

13 days ago

The capacity of a storage battery, such as those used in automobile electrical systems, is rated in ampere-hours (A?h). A 50 A?h

battery can supply a current of 50 Afor 1.0 h, or 25 A for 2.0 h, and so on.A) What total energy can be supplied by a 13V , 60A?h battery if its internal resistance is negligible?Answer= ...... JB) What volume (in liters) of gasoline has a total heat of combustion equal to the energy obtained in part (a)? (See Section 17.6; the density of gasoline is 90 kg/m 3.)Answer= ........ LC) If a generator with an average electrical power output of 0.45 kW is connected to the battery, how much time will be required for it to charge the battery fully?Answer= ........ h

Physics

You might be interested in

If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2

Maru [3345]

Answer:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

Consequently, the temperature difference across the material will be $\Delta T = 375 K$

Explanation:

In this case, we apply the Fourier Law of heat conduction expressed by the following equation:

$Q = -kA \frac{\Delta T}{\Delta x}$ (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 denotes the cross-sectional area

Q= 3KW signifies the heat transfer rate

$\Delta T$ is the temperature difference we need to determine

represents the thickness of the material $\Delta x=2.5 cm =0.025 m$

To isolate $\Delta T$ from equation (1), we obtain:

$\Delta T =\frac{Q \Delta x}{Ak}$

Initially, we convert 3KW to W, resulting in:

$Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W$

With all variables accounted for, we can substitute and calculate:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

Thus, the temperature difference across the material will be $\Delta T = 375 K$

5 0

1 month ago

Joule’s law is a linear relationship, that is, the more heat you provide, the greater the temperature change. However, in the pr

inna [3103]

the term is hydrogen bridge.

7 0

1 month ago

A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0.0

serg [3582]

Since you've completed parts a and b, I will tackle part c.
For part C
To respond to this question, we must identify the zeros of the velocity function:
$v(t)=0.04t^3-0.06t^2$
This polynomial can be factored:
$v(t)=0.04t^3-0.06t^2=t^2(0.04t-0.06)$
Finding the zeros now becomes straightforward since the function equals zero when any factor is zero.
$t^2=0;\\ 0.04t-0.06=0$
By solving these equations, we identify our zeros:
$t_1=0; t_2=\frac{3}{2}$
The particle remains stationary at t=0 and t=3/2.
For part D
We must discover when the velocity function exceeds zero. We will utilize its factored form.
We will assess when each factor is greater than zero and compile the findings in the following table:
$\centering \label{my-label} \begin{tabular}{lllll} Range & -\infty & 0 & 3/2 & +\infty \\ t^2 & - & + & + & + \\ 0.04t-0.06 & - & - & + & + \\ t^2 (0.04t-0.06) & + & - & + & + \end{tabular}$
From the table, it's evident that our function is positive when $- \infty < t$ and t>3/2.
This indicates the interval during which the particle moves forward.
For part E
The distance traveled can be represented as:
$s(t)=0.01t^4 - 0.02t^3$
We simply substitute t=12 to calculate the total distance traveled:
$s(12)=0.01(12)^4 - 0.02(12)^3=172.80 ft$
For part F
Acceleration is defined as the rate at which velocity changes.
We determine acceleration by deriving the velocity function concerning time.
$a(t)=\frac{dv}{dt}=(0.04t^3-0.06t^2)'=0.12t^2-0.12t$
To find the acceleration at 1 second, we substitute t=1s into the previous equation:
$a(1)=0.12-0.12=0$

7 0

2 months ago

A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coeffi

Ostrovityanka [3204]

Response:

$\text{heat loss} = 24864.05 \ W/m^2$

Clarification:

$T_1$ , $T_2$ represent the temperatures of gases and liquids in Kelvins,

$t_1$ and $t_2$ denote the thicknesses of the gas layer and steel slab in meters,
$h_1$ , $h_2$ are the convection coefficients for gas and liquid in $W/m^2 \cdot K$ ,
$R_c$ represents the contact resistance in $m^2 \cdot K/W$ ,