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2 years ago

Can any kind soul help me please

Physics

1 answer:

Alex2 years ago

3 0

Answer:

I) 420000J

ii)

Explanation:

(I) so you can use the formula for quantity of heat then substitute the values given

formula-Q=mc∆9

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5. the combustion of a single molecule of methane produces about 10 ev of energy. a methane molecule has a mass of 16 amu. the f

Alchen [17]

The mass of methane will produce as much energy as a single gram of uranium-235 1,287 kilograms of methane. Option D is correct.

<h3>What is uranium?</h3>

Uranium, with the atomic number 92 and the symbol U in the periodic table, is a weakly radioactive element. One of the heavy metals that may be used as a concentrated energy source is uranium.

Given data;

⇒One mole of U-235 = 235 grams

=> 1 gram of U-235 = 1/235 moles

⇒1 mole of U-235 = 6.023 x 10²³ atoms

The no of atoms is;

=> 1/235 moles of U-235

⇒N = 6.023 x 10²³/235 atoms

⇒N=25.6 x 10²⁰ atoms

If one atom of U-235 gives is 189 x 10⁶ eV energy,25.6 x 10²⁰ atoms of U-235 gives;

=> 25.6 x 10²⁰ atoms of U-235 gives;

E = 25.6 x 10²⁰ x 189 x 10⁶

E= 4.8 x 10²⁹ eV energy

One methane molecule produces;

E = 10 eV of energy

=> To produce 4.8 x 10²⁹ eV energy, no. of molecules required;

= 4.8 x 10²⁸

6.032 x 10²³ molecules of methane = 16 gms

=> 4.8 x 10²⁸ molecules of methane = 16 x 4.8 x 10²⁸/ 6.032 x 10²³ gms

m= 12.75 x 10⁵ gms

m= 1275 kilograms of methane

m≅ 1287 kilograms of methane

1 gram of U-235 has the same amount of energy as 1287 kilograms of methane.

Hence option D is correct,

To learn more about uranium refer to the link;

brainly.com/question/9099776

#SPJ1

7 0

1 year ago

A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that i

mars1129 [50]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The electric field at that point is $E = 7500 \ N/C$

Explanation:

From the question we are told that

The radius of the inner circle is $r_i = 0.80 \ m$

The radius of the outer circle is $r_o = 1.20 \ m$

The charge on the spherical shell $q_n = -500nC = -500*10^{-9} \ C$

The magnitude of the point charge at the center is $q_c = + 300 nC = + 300 * 10^{-9} \ C$

The position we are considering is x = 0.60 m from the center

Generally the electric field at the distance x = 0.60 m from the center is mathematically represented as

$E = \frac{k * q_c }{x^2}$

substituting values

$E = \frac{k * q_c }{x^2}$

where k is the coulomb constant with value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

substituting values

$E = \frac{9*10^9 * 300 *10^{-9}}{0.6^2}$

$E = 7500 \ N/C$

7 0

3 years ago

Melting wax

MaRussiya [10]

It would have to be c because it is a chemical change. your welcome!!!!!

3 0

3 years ago

Read 2 more answers

A pulley system has an efficiency of 74.2%. If you perform 200 J of work, how

Fudgin [204]

Answer:

Explanation:

If a pulley system has an efficiency of 74.2%, then only that fraction of the work performed will be useful. 74.2%=0.742. 0.742*200 is about 148J. Hope this helps!

8 0

2 years ago

Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

3 years ago

Can any kind soul help me please​

Can any kind soul help me please