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2 years ago

Why is energy transferred from the substance to the surroundings when a substance freezes

Physics

1 answer:

jeyben [28]2 years ago

7 0

Zalice sucks chakra woman trAnsfer to the fezze y try e

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Use the given data to calculate the total mass of hydrogen available for fusion over the lifetime of the sun.

pogonyaev

Total mass of the Sun = 2x10^30kg

<span>So 76% of that = (2x10^30kg)*(0.76) = 1.52x10^30kg ----> total amount of Hydrogen i</span><span>f only 12% of that is used for fusion, then (1.52x10^30kg)*(0.12) = 1.82x10^9kg</span>

8 0

2 years ago

A 1.50 µF capacitor and a 3.50 µF capacitor are connected in series across a 2.50 V battery. How much charge (in µC) is stored o

Nataly_w [17]

Explanation:

The given data is as follows.

$C_{1} = 1.50 \times 10^{-6} F$

$C_{1} = 3.50 \times 10^{-6} F$

Voltage = 2.50 V

Hence, calculate the equivalence capacitor as follows.

$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$

$\frac{1}{C} = \frac{1}{1.50 \times 10^{-6} F} + \frac{1}{3.50 \times 10^{-6} F}$

= $0.945 \times 10^{-6} F$

C = $1.06 \times 10^{-6} F$

Now, we will calculate the charge across each capacitance as follows.

Q = CV

= $1.06 \times 10^{-6} F \times 2.50 V$

= $2.65 \times 10^{-6} C$

= $2.65 \mu C$

Thus, we can conclude that $2.65 \mu C$ is the charge stored on each given capacitor.

5 0

2 years ago

A wave is sent along the first rope transmitting a power of 57.3 W. It has a wavelength of 5.54 cm and velocity of 13.87 m/s The

Serhud [2]

Answer:

$A = 2.43*10^{-3} m$

Explanation:

power through string can be determined as shown in figure

$P = 2\pi ^2 HVA^2F^2$

Where

P = 57.3 W

V = 13.87 m/s

H = 567 g/m

we know that

$V = f *\lambda$ $\lambda = \frac{v}{f}$

therefore $P = 2\pi ^2 HVA^2(\frac{v}{f})^2$

$57.3 = \frac{2\pi ^2 * 0.567 *13.87^{3}* A^2}{(5.54*10^{-2})^2}$

$A = 2.43*10^{-3} m$

7 0

2 years ago

A spherical drop of water carrying a charge of 43 pC has a potential of 540 V at its surface (with V = 0 at infinity). (a) What

almond37 [142]

Explanation:

Given that,

Charge on a spherical drop of water is 43 pC

The potential at its surface is 540 V

(a) The electric potential on the surface is given by :

$V=\dfrac{kq}{r}$

r is the radius of the drop

$r=\dfrac{kq}{V}\\\\r=\dfrac{9\times 10^9\times 43\times 10^{-12}}{540}\\\\r=7.17\times 10^{-4}\ m$

(b) Let R is the radius of the spherical drop, when two such drops of the same charge and radius combine to form a single spherical drop. ATQ,

$\dfrac{4}{3}\pi r^3+\dfrac{4}{3}\pi r^3=\dfrac{4}{3}\pi R^3\\\\2r^3=R^3\\\\R=2^{1/3} r$

Now the charge on the new drop is 2q. New potential is given by :

$V=\dfrac{9\times 10^9\times 43\times 10^{-12}\times 2}{2^{1/3}\times 7.17\times 10^{-4}}\\\\V=856.79\ V$

Hence, the radius of the drop is $7.17\times 10^{-4}\ m$ and the potential at the surface of the new drop is 856.79 V.

5 0

2 years ago

How do scientists determine motion of objects in the universe??

Katen [24]

Usually, Scientists use "Parallax Method" in order to determine the different positions of an object

Hope this helps!

5 0

2 years ago

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