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3 years ago

You should check your battery ___________.

Physics

2 answers:

NNADVOKAT [17]3 years ago

8 0

The battery should be checked every six month to ensure that it is working properly.

Explanation:

The battery is the power source for the car as it provides power for several functions that are going in the car. The battery used in a car is always in use whenever the car is driven.

The battery used in a car should be checked for any malfunctioning as it may cause harm to it. The battery may be dead earlier than its life if it is not been taken care at regular intervals.

So, in order to keep the battery of the car working properly, it should be ensured that the battery is checked every five-six month or at least twice in a year. This way, if there is any malfunctioning in any part of it, it can be detected and can be made correct in appropriate time.

Thus, The battery should be checked every six month to ensure that it is working properly.

Learn More:

1. Compare the types of friction brainly.com/question/1800116

2. Fossil which is formed when the outline of original organism is formed brainly.com/question/2856442

3. Why does the refraction of light takes place brainly.com/question/3095091

Answer Details:

Grade: Middle School

Subject: Physics

Chapter: Battery

Keywords:

Every six month, every week, car, battery, malfunctioning, detected, appropriate, power source, driving.

podryga [215]3 years ago

5 0

One should check the battery every week.

A battery if not maintained properly can cause it to die earlier or get damaged. The reasons why should the battery be checked regularly is that due to improper management of the battery and also the weather conditions, the battery usually gets corroded easily and the battery won't be able to work properly.

The connections must be checked properly so that the battery gets charged properly and should not die while driving.

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The earth____ around the sun because the earth___ we have night and day

Harrizon [31]

Answer:

The Earth Revolves around the sun because the Earth Tilts, we have night and day.

Explanation:

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4 0

3 years ago

The resistance between 2 points in an electrical circuit is 1.1 Ω. What additional resistance

valentinak56 [21]

Answer:

the new resister is 11 ohms.

Explanation:

Set it up like this.

1/x + 1/1.1 = 1 Subtract 1/1.1 from both sides

1/x = 1 - 1/1.1

1 - 1/1.1 = 1/11

1/x = 1/11 Cross multiply

11 = x

If 1/11 bothers you, you could do it it another way.

1 - 1/1.1 = (1.1 - 1 ) / 1.1 = 0.1 / 1.1 Multiply top and bottom by 10

0.1*10/(1.1 * 10 ) = 1 / 11

5 0

3 years ago

ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami

yuradex [85]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

Explanation :

The given balanced chemical reaction is,

$2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{product}$

$\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]$

$\Delta H^o=62.2kJ=62200J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{product}$

$\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]$

$\Delta S^o=132.67J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 298 K.

$\Delta G^o=(62200J)-(298K\times 132.67J/K)$

$\Delta G^o=22664.34J=22.66kJ$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

6 0

3 years ago

(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50

Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, $p=m_Av_A=m_Bv_B=m_Cv_C$ , were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as $m_h$ , and the mass of the first and second coals as $m_1$ and $m_2$ respectively

We start with the transition between parts A and B, so we have:

$m_Av_A=m_Bv_B$

Which means

$m_hv_A=(m_h+m_1)v_B$

And since we want the mass of the first coal thrown ( $m_1$ ) we do:

$m_hv_A=m_hv_B+m_1v_B$

$m_hv_A-m_hv_B=m_1v_B$

$m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}$

Substituting values we obtain

$m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg$

For the transition between parts B and C, we can write:

$m_Bv_B=m_Cv_C$

Which means

$(m_h+m_1)v_B=(m_h+m_1+m_2)v_C$

Since we want the new final speed of the car ( $v_C$ ) we do:

$v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}$

Substituting values we obtain

$v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s$

5 0

3 years ago

F = 2.0*10^20 N,

natka813 [3]

$F=\dfrac{Gm_1m_2}{r^2}$

With the given values of $F,G,m_1,r$ , we have

$2.0\times10^{20}\,\mathrm N=\dfrac{\left(6.67\times10^{-11}\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98\times10^{24}\,\mathrm{kg}\right)m_2}{\left(3.8\times10^8\,\mathrm m\right)^2}$