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3 years ago

14

How long will it take to travel 2000 Km at 500 Km a.100,000 hr b.2500 Km c.4 hr

1 answer:

IRISSAK [1]3 years ago

5 0

Answer:

the question seems incomplete but the most reasonable answer is 4 times

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4 years ago

A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing throu

Setler79 [48]

<h2>Given :</h2>

total charge = 9.0 mC = 0.009 C

Each electron has a charge of :

$1.6 \times 10 {}^{ - 19} \: C$

For producing 1 Cuolomb charge we need :

$\mathrm{\dfrac{1}{1.6 \times 10 {}^{ - 19} } }$

$\dfrac{10 {}^{19} }{1.6}$

$\dfrac{10\times 10 {}^{19} }{16}$

$\dfrac{100 \times 10 {}^{18} }{16}$

$\mathrm{6.24 \times 10 {}^{18} \: \: electrons}$

Now, for producing 0.009 C of charge, the number of electrons required is :

$0.009 \times 6.24 \times {10}^{18}$

$0.05616 \times 10 {}^{18}$

$\mathrm{5.616 \times 10 {}^{16} \: \: electons}$

_____________________________

So, Number of electrons passing through the cross section in 3.6 seconds is :

$\mathrm{5.616 \times 10 {}^{16} \: \: electrons}$

Number of electrons passing through it in 1 Second is :

$\dfrac{5.616 \times {10}^{16} }{3.6}$

$\mathrm{1.56 \times 10 {}^{16} \: \: electrons}$

Now, in 10 seconds the number of electrons passing through it is :

$10 \times \mathrm{1.56 \times 10 {}^{16} \: \: }$

$\mathrm{1.56 \times 10 {}^{17} \: \: electrons}$

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$\mathrm{ \#TeeNForeveR}$

6 0

3 years ago

How much heat h1 is transferred to the skin by 25.0 g of steam onto the skin? the latent heat of vaporization for steam is l=2.2

elena-s [515]

The heat transferred by the steam to the skin is given by
$Q=m L_v$
where
m is the mass of the steam
$L_v$ is the latent heat of vaporization.

In our problem, the mass of the steam is (converting into kg)
$m=25.0 g=0.025 kg$
while the latent heat of vaporization of the steam is
$L_v = 2.256 \cdot 10^6 J/kg$
Substituting into the previous formula, we find the heat transferred to the skin:
$Q=m L_v = (0.025 kg)(2.256 \cdot 10^6 J/kg)=56400 J = 2.56 \cdot 10^4 J$

8 0

3 years ago