█ Question <span>█
</span><span>In an electronic transition, an atom cannot emit what?
</span>█ Answer █
When an electronic transition is occurring, an atom cannot emit ultra-violet light.
<span>Hope that helps! ★ <span>If you have further questions about this question or need more help, feel free to comment below or leave me a PM. -UnicornFudge aka Nadia</span></span>
Answer:
Explanation:
mass of 1 L water = 1 kg .
200⁰F = (200 - 32) x 5 / 9 = 93.33⁰C .
260.928 K = 260.928 - 273 = - 12.072⁰C .
water is at higher temperature .
Let the equilibrium temperature be t .
Heat lost by water = mass x specific heat x fall of temperature
= 1 x 4.2 x 10³ x ( 93.33 - t )
Heat gained by copper
= .25 x .385 x 10³ x ( t + 12.072 )
Heat lost = heat gained
1 x 4.2 x 10³ x ( 93.33 - t ) = .25 x .385 x 10³ x ( t + 12.072 )
93.33 - t = .0229 ( t + 12.072)
93.33 - t = .0229 t + .276
93.054 = 1.0229 t
t = 90.97⁰C .
Answer:
recall that heat absorbed released is given by
Q = mc*(T2 - T1)
where
m = mass (in g)
c = specific heat capacity (in J/g-k)
T = temperature (in C or K)
*note: Q is (+) when heat is absorbed and (-) when heat is released.
substituting,
Q = (480)*(0.97)*(234 - 22)
Q = 98707 J = 98.7 kJ
Explanation:
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
Answer:
A) 
= 1.44 kg m², B) moment of inertia must increase
Explanation:
The moment of inertia is defined by
I = ∫ r² dm
For figures with symmetry it is tabulated, in the case of a cylinder the moment of inertia with respect to a vertical axis is
I = ½ m R²
A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is
I = 
+ m D²
Let's apply these equations to our case
The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms
=
+ 2 
= ½ M R²
The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body
M = 7/8 m total
M = 7/8 64
M = 56 kg
The mass of the arms is
m’= 1/8 m total
m’= 1/8 64
m’= 8 kg
As it has two arms the mass of each arm is half
m = ½ m ’
m = 4 kg
The arms are very thin, we will approximate them as a particle
= M D²
Let's write the equation
= ½ M R² + 2 (m D²)
Let's calculate
= ½ 56 0.20² + 2 4 0.20²
= 1.12 + 0.32
= 1.44 kg m²
b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase
