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3 years ago

6

Exit to create electricity using a nuclear power plant, energy from the uranium in the reactor produces lots of ________ in orde

r to make the generator work.
a. fuel
b. heat
c. water
d. electricity

1 answer:

inessss [21]3 years ago

8 0

Option b is ur answer

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How to find initial velocity in projectile motion problems when you are not given the vlocity?

Papessa [141]

Vi= square root vi^2 -2ad

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3 years ago

Your car's 30.0 W headlight and 2.50 kW starter are ordinarily connected in parallel in a 12.0 V system. What power (in W) would

Tatiana [17]

Answer:

<h3>The power of headlight in series connection is 29.64 W</h3>

Explanation:

Given :

Power of headlight $P_{1} = 30$ W

Power of starter $P_{2} = 2500$ W

Voltage of headlight and starter $V = 12$ V

From equation of power,

$P = \frac{V^{2} }{R}$

$R = \frac{V^{2} }{P}$

For finding the resistance of headlight and starter,

⇒ For headlight,

$R_{1} = \frac{144}{30} = 4.8$ Ω

⇒ For starter,

$R_{2} = \frac{144}{2500} = 0.057$ Ω

Since equivalent resistance,

$R_{eq} = R_{1} + R_{2} + ........$

$R_{eq} = 4.8 +0.057 = 4.857$ Ω

So power in series is given by,

$P_{s } = \frac{V^{2} }{R_{eq} } = \frac{144}{4.857}$

$P_{s } = 29.64$ W

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3 years ago

Please help me with this 29 points

Irina18 [472]

Answer:

)Give the definition of poverty line as defined by the World Bank.

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3 years ago

A laser beam is incident at an angle of 30.0° from the vertical onto a solution of corn syrup in water. The beam is refracted to

dimaraw [331]

Answer with Explanation:

We are given that

Angle of incidence, $i=30^{\circ}$

Angle of refraction, $r=19.24^{\circ}$

a.Refractive index of air, $n_1=1$

We know that

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$n_2=\frac{n_1sin i}{sin r}=\frac{sin30}{sin19.24}=1.517$

b.Wavelength of red light in vacuum, $\lambda=632.8nm=632.8\times 10^{-9} m$

$1nm=10^{-9} m$

Wavelength in the solution, $\lambda'=\frac{\lambda}{n_2}$

$\lambda'=\frac{632.8}{1.517}=417nm$

c.Frequency does not change .It remains same in vacuum and solution.

Frequency, $\nu=\frac{c}{\lamda}=\frac{3\times 10^8}{632.8\times 10^{-9}}$

Where $c=3\times 10^8 m/s$

Frequency, $\nu=4.74\times 10^{14}Hz$

d.Speed in the solution, $v=\frac{c}{n_2}$

$v=\frac{3\times 10^8}{1.517}=1.98\times 10^8m/s$

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3 years ago

Write the equation that links current, potential difference, and resistance

mrs_skeptik [129]

You can write the equation in 3 different ways, depending on which quantity you want to be the dependent variable. Any one of the three forms can be derived from either of the other two with a simple algebra operation. They're all the same relationship, described by "Ohm's Law".

==> Current = (potential difference) / (resistance)

==> Potential difference = (current) x (resistance)

==> Resistance = (potential difference) / (resistance)

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3 years ago