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2 years ago

Suppose a log's mass is 5 kg. After burning, the mass of the ash is 1 kg. explain what May have happened to the other 4 kg

1 answer:

4 0

I think the "missing mass" may have escaped the site of the
conflagration in the form of hot, invisible gases, and in the form
of all the tiny particles in the smoke.

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Temperature is a measure of _ of the particles in an object

djverab [1.8K]

energy is the correct answer to fill the blank bb :)

3 0

2 years ago

A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. A pie

liubo4ka [24]

Answer:

Explanation:

Capacitance of the capacitor = 13.5μF

Voltage across plate is 24V

Dielectric constant k=3.55.

a. Energy in capacitor is given by

E=1/2CV^2

We want to calculate energy without the dielectric substance

Given that C=13.5 μF and V=24V

The capacitance give is with dielectric so we need to remove it

C=kCo

Co=C/k

Then the Co=13.5μF/3.55

Co=3.803μF

Then

E=(1/2)×3.803×10^-6×24^2

E=1.1×10^-3J

E=1.1mJ

b. Energy in capacitor is given by

E=1/2CV^2

The capacitance given is with a dielectric, so we are going to apply it direct.

Given that C=13.5 μF and V=24V

Then

E=(1/2)×13.5×10^-6×24^2

E=3.89×10^-3J

E=3.9mJ

c. The energy without dielectric is 1.1mJ and the energy with dielectric is 3.9mJ

The energy increase when the dielectric material is added

d. Dielectrics in capacitors serve three purposes: to keep the conducting plates from coming in contact, allowing for smaller plate separations and therefore higher capacitances;

Therefore, Since dielectric allow higher capacitance, and energy of a capacitor is directly proportional to the capacitance, then the higher the capacitance the higher the energy.

6 0

3 years ago

HELP ASAPPPPPPP !!!!!!!!!

Misha Larkins [42]

The answer is hydrogen

3 0

3 years ago

A simple dipole consists of two charges with the same magnitude, q, but opposite sign separated by a distance d. The EDM (electr

DedPeter [7]

Answer:

a. dW = ∫pEsinθdθ b. W = p.E

Explanation:

a. We know torque τ = p × E = pEsinθ where θ is the angle between p and E

Let the torque τ rotate the dipole by an amount dθ. So, the workdone dW = ∫τdθ = ∫pEsinθdθ

b. So, the total work done is gotten by integrating from 90 to θ. So,

W = ∫₉₀⁰dW

= ∫₉₀⁰pEsinθdθ

= pE∫₉₀⁰sinθdθ

= pE(cosθ - cos90)

=pEcosθ

= p.E

8 0

3 years ago

n ultraviolet light beam having a wavelength of 130 nm is incident on a molybdenum surface with a work function of 4.2 eV. How f

pashok25 [27]

Answer:

The speed of the electron is 1.371 x 10⁶ m/s.

Explanation:

Given;

wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m

the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J

The energy of the incident light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f = c / λ

$E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{130*10^{-9}} \\\\E = 15.291*10^{-19} \ J$

Photo electric effect equation is given by;

E = W₀ + K.E

Where;

K.E is the kinetic energy of the emitted electron

K.E = E - W₀

K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J

K.E = 8.563 x 10⁻¹⁹ J

Kinetic energy of the emitted electron is given by;

K.E = ¹/₂mv²

where;

m is mass of the electron = 9.11 x 10⁻³¹ kg

v is the speed of the electron

$v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2*8.563*10^{-19}}{9.11*10^{-31}}}\\\\v = 1.371 *10^{6} \ m/s$

Therefore, the speed of the electron is 1.371 x 10⁶ m/s.

8 0

3 years ago