The height of the oil column above the water in the vessel is determined as 2 cm.
<h3>
Pressure of the vessel</h3>
The pressure of the vessel due to water, oil and silver poured into the vessel is determined from mercury column.
let level of mercury = 20 cm + 0.5 cm = 20.5 cm
20.5 cmHg = 205 mmHg
1 mmHg = 133.32 Pa
205 mmHg = 27,330.6 Pa
<h3>Height of the liquids in the vessel</h3>
P = ρgh
where;
ρ is the density of water, oil and silver respectively
ρ = 1000 kg/m³ + 881 kg/m³ + 10,800 kg/m³ = 12,681 kg/m³
h = P/(ρg)
h = (27,330.6) / (12,681 x 9.8)
h = 0.22 m
h = 22 cm
<h3>Height of oil column</h3>
Oil is less dense than water and will float on water.
Height of oil column = 22 cm - 20 cm = 2 cm
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Answer:
E₄ = - 0.85 eV
E₂ = - 3.4 eV
Ephoton = 2.55 eV
Explanation:
The sum of Kinetic Energy (K) and Potential Energy (U) of the Helium atom is equal to the total energy of Helium atom in the specified state N. From Bohr's atomic model, the energy of a hydrogen atom in state N is given as:
En = K + U = (-1/n²)(13.6 eV)
a)
Here,
n = 4
Therefore,
E₄ = (-1/4²)(13.6 eV)
<u>E₄ = - 0.85 eV</u>
<u></u>
b)
Here,
n = 2
Therefore,
E₂ = (-1/2²)(13.6 eV)
<u>E₂ = - 3.4 eV</u>
<u></u>
c)
The energy of photon emitted in the transition from level 4 to level 2 will be equal to the difference in the energy of both levels:
Ephoton = ΔE = E₄ - E₂
Ephoton = - 0.85 eV - (- 3.4 eV)
<u>Ephoton = 2.55 eV</u>
<u></u>
The complete question is;
James Joule (after whom the unit of energy is named) claimed that the water at the bottom of Niagara Falls should be warmer than the water at the top, 51 m above the bottom. He reasoned that the falling water would transform its gravitational potential energy at the top into thermal energy at the bottom, where turbulence brings the water almost to a halt. If this transformation is the only process occurring, how much warmer will the water at the bottom be?
Answer:
Water becomes warmer by a temperature of ΔT = 0.119 K
Explanation:
If we assume that gravitational kinetic energy will be converyrf into thermal enrgy, we will have;
Q = U
So, m•c_w•ΔT = mgh
Where;
c_w is specific heat capacity of water with a value of 4184 J/Kg.K
ΔT is change in temperature indicating how warmer the water will be. Thus making ΔT the subject, we have;
ΔT = gh/c_w
So, ΔT = 9.8 x 51/4184 = 0.119 K
First of all,
The electric field at the surface of the sphere is given by
E = kQ/r²
The field strength at which breakdown occurs in the air is <span>3.0 MV/m
</span>
So, E = 3.0 MV/m
<span>The sphere potential is defined as
V = kQ/r</span>
<span>If we divide E/V we get
E = V/r </span>
<span>r = V/E = 20000V / 3.10^6 V/m = 6.66 </span>10^-3 m = 6.66 mm
2. Charge
<span>V = kQ/r .............>>
</span>
Q = Vr/k = 20000V *( 6.66 10^-3 m)/ (9.10^9 N m2/C2) = 1.481 10^-8 C
The angle of inclination is calculated using sin
function,
sin θ = 5 m / 20 m = 0.25
θ = 14.4775°
<span>The net force exerted is then calculated:
F net = m g sin θ = 20 * 9.8 * 0.25 </span>
F net = 49N
<span>Work is product of net force and distance:
W = F net * d = 49 * 20 </span>
<span>Work = 980 J </span>