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3 years ago

WHY ADIABAT IS STEEPER THAN ISOTHERM???

Physics

1 answer:

Anna35 [415]3 years ago

6 0

Bruh I’m honestly just going to take the free points

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Explain why sedimentary rocks are found as a veneer covering large areas of the continental igneous rocks

musickatia [10]

Sedimentary rocks are the result of igneous rocks<span> breaking down.This is a slow process so there is not as much broken down as there is not broken down.</span>

6 0

3 years ago

Read 2 more answers

When operated on a household 110.0 V line, typical hair dryers draw about 1650 W of power. The current can be modeled as a long,

Andre45 [30]

Explanation:

Given that,

Voltage of household line, V = 110 V

Power of the hairdryer, P = 1650 W

During use, the current is about 1.95 cm from the user's hand.

(a) Power is given by :

$P=V\times I\\\\I=\dfrac{P}{V}\\\\I=\dfrac{1650\ W}{110\ V}\\\\I=15\ A$

(b) Again the power is given by :

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

R is resistance of the dryer

$R=\dfrac{V^2}{P}\\\\R=\dfrac{(110)^2}{1650}\\\\R=7.34\ \Omega$

(c) The magnetic field produced by the dryer at the user's hand is given by :

$B=\dfrac{\mu_o I}{2\pi r}\\\\B=\dfrac{4\pi \times 10^{-7}\times 15}{2\pi \times 1.95\times 10^{-2}}\\\\B=1.53\times 10^{-4}\ T$

Hence, this is the required solution.

4 0

2 years ago

a fast humvee drove from desert a to desert b. for the 12 hours, it travelled at an speed of 185 km/h. for the next 13 hours, it

ivanzaharov [21]

Answer:

I am not sure of the question?

Explanation:

List the question

8 0

3 years ago

A 2. 0 μf and a 4. 0 μf capacitor are connected in series across an 8. 0-v dc source. what is the charge on the 2. 0 μf capacito

Nezavi [6.7K]

voltage across 2.0μf capacitor is 5.32v

Given:

C1=2.0μf

C2=4.0μf

since two capacitors are in series there equivalent capacitance will be

[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]

$c = \frac{c1 \times c2}{c1 + c2}$

$= \frac{2 \times 4}{2 + 4}$

=1.33μf

As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.

Q=CV

given,V=8v

$= 1.33 \times 10 {}^{ - 6} \times 8$