Ask question

Ask question

All categories

2 years ago

5

A particular type of automobile storage battery is characterized as "207 - Ampere - hour, 9.4 V."

1 answer:

zepelin [54]2 years ago

6 0

Electric Energy = Current x Voltage x time = 207 x 9.4 = 1,945.8 J

You might be interested in

How many electrons are in the element that comes just before platinum in the periodic table?

laila [671]

The element is iridium and it has 77 electrons

7 0

3 years ago

PLZZZZZ I NEED HELP Which is an adaptation that helps birds maintain a stable body temperature? A. air sacs connected to lungs B

aniked [119]

Answer:

c

Explanation:

trust

5 0

3 years ago

Read 2 more answers

What is the purpose of a rearview mirror in a car?

Shalnov [3]

The third choice.

The driver wants to see the object that is behind him. The light reflects off the mirror into the eyes of the driver portraying the object behind him

5 0

2 years ago

Read 3 more answers

What is the energy (in evev) of a photon of visible light that has a wavelength of 500 nmnm?

lisabon 2012 [21]

<h3>Answer:</h3>

E≈2,5 eV

<h3>Explanation:</h3>

_______________

λ=500 nm = 500·10⁻⁹ m

c=3·10⁸ m/s

h=6,63·10⁻³⁴ J·s = 4,14·10⁻¹⁵ eV·s

_______________

E - ?

_______________

$\displaystyle \boldsymbol{E}=h\nu =h \frac{c}{\lambda} =4,14\cdot 10^{-15} \; eV\cdot s\cdot \; \frac{3\cdot 10^8\; m/s}{500\cdot 10^{-9}\; m} =2,484\; eV\approx \boldsymbol{2,5\; eV}$

6 0

1 year ago

A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until t

postnew [5]

Answer:

Explanation:

Given

mass of saturated liquid water $m=1.4\ kg$

at $200^{\circ}$ specific volume is $\nu =0.001157\ m^3\kg$ (From Table A-4,Saturated water Temperature table)

$V_1=m\nu _1$

$V_1=1.4\times 0.001157$

$V_1=1.6198\times 10^{-3}\ m^3$

Final Volume $V_2=4V_1$

$V_2=4\times (1.6198\times 10^{-3})$

$V_2=6.4792\times 10^{-3}\ m^3$

Specific volume at this stage

$\nu _2=\frac{V_2}{m}$

$\nu _2=\frac{6.4792\times 10^{-3}}{1.4}$

$\nu _2=0.004628\ m^3/kg$

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

$T_2=T_1^{*}+\frac{T_2^{*}-T_1^{*}}{\alpha _2^{*}-\alpha _1^{*}}\times (\alpha _2-\alpha _1^{*})$

$T_2=370^{\circ}+\frac{373.95-370}{0.003106-0.004953}\times (0.004628-0.004953)$

$T_2=370.7^{\circ} C$

4 0

3 years ago