Ask question

Ask question

All categories

2 years ago

14

A car battery seems to be malfunctioning (not providing the proper voltage and current to start your car). While the car is off

and you are not trying to start it, a friend uses a voltmeter to measure the voltage and finds that your battery voltage is 12.0 V, just as it is supposed to be. After the voltage check, you try to start your car again to no avail. Can you rule out the battery as the problem?

1 answer:

Stella [2.4K]2 years ago

4 0

No. You cannot rule out the battery even after the open circuit voltage measurement. The open-circuit voltage may not have changed but the battery's internal resistance may have greatly increased.

You might be interested in

Interacting with others while exercising is beneficial to social health because it allows a person to __________.

vlada-n [284]

Feel better and develop communication skills

It's not only the physical well-being that has developed as well as intellectual and emotional aspect of an individual. When having conversation to someone and you are doing something it's also the same of having a multitask work. That all senses response quickly and something is developing in you, at same time you are establishing good rapport towards others.

6 0

2 years ago

One normal afternoon in undisclosed City X, the chocolate factory workers overload the Easter egg machine. A group of angsty tee

musickatia [10]

Answer:

The correct option is;

C. 1,715 m

Explanation:

We are given the information from the group of teen at the City edge

Time of arrival of explosion sound = 5 s after sighting

Time of sighting explosion = 5 s before hearing the boom

Speed of sound in air ≈ 343 m/s

Speed of light = 299,792 km/s

Therefore, distance covered by sound in 5 seconds is given by the following equation;

$Speed = \frac{Distance}{Time}$

$\therefore 343 \ m/s= \frac{Distance}{5 \, s}$

Hence Distance = 343 m/s × 5 s = 1715 m

To check, we compare the time it would take for the light to cover 1715 m

That is $Time = \frac{Distance}{Speed} = \frac{1715}{299,792,000} = 0.00000572 \, s$ which is instantaneous hence the distance can be approximated by the time duration for the speed of sound.

Therefore, the distance of the students from the factory is approximately 1,715 m

8 0

3 years ago

Answer the following question according to formal grammatical rules.

VARVARA [1.3K]

Answer:

c

Explanation:

just trying to follow basic grammar.

5 0

2 years ago

An object in space is initially stationary relative to the Earth. Then, a force begins acting on the object, starting with a for

Thepotemich [5.8K]

Answer:

21741 s i believe

Explanation:

8 0

3 years ago

Moist air initially at 1258C, 4 bar, and 50% relative humidity is contained in a 2.5-m3 closed, rigid tank. The tank contents ar

brilliants [131]

Here is the missing part of the question

To Determine the heat transfer, in kJ if the final temperature in the tank is 110 deg C

Answer:

Explanation:

The image attached below shows the process on T - v diagram

<u>At State 1:</u>

The first step is to find the vapor pressure

$P_{v1} = \rho_1 P_g_1$

$= \phi_1 P_{x \ at \ 125^0C}$

= 0.5 × 232 kPa

= 116 kPa

The initial specific volume of the vapor is:

$P_{v_1} v_{v_1} = \dfrac{\overline R}{M_v}T_1$

$116 \times 10^3 \times v_{v_1} = \dfrac{8314}{18} \times (125 + 273)$

$116 \times 10^3 \times v_{v_1} = 183831.7778$

$v_{v_1} = 1.584 \ m^3/kg$

<u>At State 1:</u>

The next step is to determine the mass of water vapor pressure.

$m_{v1} = \dfrac{V}{v_{v1}}$

$= \dfrac{2.5}{1.584}$

= 1.578 kg

Using the ideal gas equation to estimate the mass of the dry air $m_a$ $P_{a1} V = m_a \dfrac{\overline R}{M_a}T_1$

$(P_1-P_{v1}) V = m_a \dfrac{\overline R}{M_a}T_1$

$(4-1.16) \times 10^5 \times 2.5 = m_a \dfrac{8314}{28.97}\times ( 125 + 273)$

$710000= m_a \times 114220.642$

$m_a = \dfrac{710000}{114220.642}$

$m_a = 6.216 \ kg$

For the specific volume $v_{v_1} = 1.584 \ m^3/kg$ , we get the identical value of saturation temperature

$T_{sat} = 100 + (110 -100) \bigg(\dfrac{1.584-1.673}{1.210 - 1.673}\bigg)$

$T_{sat} =101.92 ^0\ C$

Thus, at $T_{sat} =101.92 ^0\ C$ , condensation needs to begin.

However, since the exit temperature tends to be higher than the saturation temperature, then there will be an absence of condensation during the process.

Heat can now be determined by using the formula

Q = ΔU + W

Recall that: For a rigid tank, W = 0

Q = ΔU + 0

Q = ΔU

Q = U₂ - U₁

Also, the mass will remain constant given that there will not be any condensation during the process from state 1 and state 2.

<u>At State 1;</u>

The internal energy is calculated as:

$U_1 = (m_a u_a \ _{ at \ 125^0 C})+ ( m_{v1} u_v \ _{ at \ 125^0 C} )$

At $T_1$ = 125° C, we obtain the specific internal energy of air

SO;

$U_{a \ at \ 125 ^0C } = 278.93 + ( 286.16 -278.93) (\dfrac{398-390}{400-390} )$

$=278.93 + ( 7.23) (\dfrac{8}{10} )$

$= 284.714 \ kJ/kg\\$

At $T_1$ = 125° C, we obtain the specific internal energy of water vapor

$U_{v1 \ at \ 125^0C} = u_g = 2534.5 \ kJ/kg$

$U_1 = (m_a u_a \ at \ _{ 125 ^0C }) + ( m_{v1} u_v \ at \ _{125^0C} )$

= 6.216 × 284.714 + 1.578 × 2534.5

= 5768.716 kJ

<u>At State 2:</u>

The internal energy is calculated as:

$U_2 = (m_a u_a \ _{ at \ 110^0 C})+ ( m_{v1} u_v \ _{ at \ 110^0 C} )$

At temperature 110° C, we obtain the specific internal energy of air

SO;

$U_{a \ at \ 110^0C } = 271.69+ ( 278.93-271.69) (\dfrac{383-380}{390-380} )$

$271.69+ (7.24) (0.3)$

$= 273.862 \ kJ/kg\\$

At temperature 110° C, we obtain the specific internal energy of water vapor

$U_{v1 \ at \ 110^0C}= 2517.9 \ kJ/kg$

$U_2 = (m_a u_a \ at \ _{ 110 ^0C }) + ( m_{v1} u_v \ at \ _{110^0C} )$

= 6.216 × 273.862 + 1.578 × 2517.9

= 5675.57 kJ

Finally, the heat transfer during the process is

Q = U₂ - U₁

Q = (5675.57 - 5768.716 ) kJ

Q = -93.146 kJ

with the negative sign, this indicates that heat is lost from the system.

6 0

2 years ago