T<span>he correct unit for electrical power is "watt".
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That<span>’s actually the unit that measures the rate per time that electric energy is transferred.</span>
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<span>Have a nice day! :)</span>
Answer:
Speed is solved with time and distance but has no direction
Average velocity is solved with Δx/Δt and has a direction
Answer:
E_particle = 1,129 10⁻²⁰ J / particle
T= 817.5 K
Explanation:
Energy is a scalar quantity so it is additive, let's look for the total energy of each gas
Gas a
E_a = 2 5000 = 10000 J
Gas b
E_b = 3 8000 = 24000 J
When the total system energy is mixed it is
E_total = E_a + E_b
E_total = 10000 + 24000 = 34000
The total mass is
M = m_a + m_b
M = 2 +3 = 5
The average energy among the entire mass is
E_averge = E_total / M
E_averago = 34000/5
E_average = 6800 J
One mole of matter has Avogadro's number of atoms 6,022 10²³ particles
Therefore, each particle has an energy of
E_particle = E_averag / 6.022 10²³ = 6800 /6.022 10²³
E_particle = 1,129 10⁻²⁰ J / particle
For find the temperature let's use equation
E = kT
T = E / k
T = 1,129 10⁻²⁰ / 1,381 10⁻²³
T = 8.175 102 K
T= 817.5 K
Answer:
Explanation:
Diffraction grating is used to form interference pattern of dark and bright band.
Distance between adjacent slits (a ) = 1 / 420 mm
= 2.38 x 10⁻³ mm
2.38 x 10⁻⁶ m
wave length of red light
= 680 x 10⁻⁹ m
For bright red band
position x on the screen
= n λD / a , n = 0,1,2,3 etc
D = distance of screen
putting n = 1 , 2 and 3 , we can get three locations of bright red band.
x₁ = λD / a
= 680 x 10⁻⁹ x 2.8 / 2.38 x 10⁻⁶
= .8 m
= 80 cm
Position of second bright band
= 2 λD / a
= 2 x 80
= 160 cm
Position of third bright band
= 3 λD / a
= 3 x 80
= 240 cm
Answer:
(a): The magnitude of the electric force on the small sphere = 
(b): Shown below.
Explanation:
<u>Given:</u>
- m = mass of the small sphere.
- q = charge on the small sphere.
- L = length of the silk fiber.
= surface charge density of the large vertical insulating sheet.
<h2>
(a):</h2>
When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:
<em>where,</em>
is the electrical permittivity of the free space.
The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:
Thus, the magnitude of the electric force on the small sphere is given by
The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.
<h2>
(b):</h2>
When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.
According to the fig.,
<em>where,</em>
= electric force on the sphere, acting along left.
= weight of the sphere, acting vertically downwards.
<em />
g is the acceleration due to gravity.
