Answer:
The balloon has more negative charges than positive charges. The tissue paper has a neutral charge.The static electricity pulls on the tissue paper lifting it up.
To solve this problem we will apply the concepts given by the principles of superposition, specifically those described by Bragg's law in constructive interference.
Mathematically this relationship is given as
Where,
d = Distance between slits
= Wavelength
n = Any integer which represent the number of repetition of the spectrum
Calculating the value for n, we have
n = 1
n=2
n =3
Therefore the intensity of light be maximum for angles 23.3° and 52.28°
Answer:
In an elastic collision, the total kinetic energy is conserved, while in an inelastic collision, it is not
Explanation:
Let's define the two types of collision:
- Elastic collision: an elastic collision is a collision in which:
1) the total momentum of the system is conserved
2) the total kinetic energy of the system is conserved
Typically, elastic collisions occur when there are no frictional forces acting on the objects in the system, so that no kinetic energy is lost into thermal energy. An example of elastic collision is the collision between biliard balls.
- Inelastic collision: an inelastic collision is a collision in which:
1 ) the total momentum of the system is conserved
2) the total kinetic energy of the system is NOT conserved
In an elastic collision, part of the total kinetic energy is lost (=converted into thermal energy) due to the presence of frictional forces. An example of inelastic collision is the accident between two cars, in which part of the energy is converted into heat.
Answer:
9704.6 J
Explanation:
Total Thermal Energy
= Energy required to bring gold to melting point + Energy required to change the state of gold from solid to liquid
= mcT + <em>m</em><em>l</em><em>f</em><em> </em><em> </em>[bolded is energy used to bring gold to melting point, <em>i</em><em>t</em><em>a</em><em>l</em><em>i</em><em>c</em><em>i</em><em>s</em><em>e</em><em>d</em><em> </em><em>i</em><em>s</em><em> </em><em>s</em><em>t</em><em>a</em><em>t</em><em>e</em><em> </em><em>c</em><em>h</em><em>a</em><em>n</em><em>g</em><em>e</em><em> </em><em>o</em><em>f</em><em> </em><em>g</em><em>o</em><em>l</em><em>d</em><em> </em><em>f</em><em>r</em><em>o</em><em>m</em><em> </em><em>s</em><em>o</em><em>l</em><em>d</em><em> </em><em>t</em><em>o</em><em> </em><em>l</em><em>i</em><em>q</em><em>u</em><em>i</em><em>d</em><em>]</em>
= (0.0500)(126)(1063 - 21) + <em>(</em><em>0</em><em>.</em><em>0</em><em>5</em><em>0</em><em>0</em><em>)</em><em>(</em><em>6</em><em>.</em><em>2</em><em>8</em><em> </em><em>×</em><em> </em><em>1</em><em>0</em><em>^</em><em>4</em><em>)</em>
= <u>9704.6</u><u> </u><u>J</u>
Answer:
r2 = 1 m
therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m
Explanation:
For this exercise we must use conservation of energy
the electric potential energy is
U = 
for the proton at x = -1 m
U₁ =
for the electron at x = 1 m
U₂ = 
starting point.
Em₀ = K + U₁ + U₂
Em₀ = 
final point
Em_f = 
energy is conserved
Em₀ = Em_f
\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})
\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²( 
)
we substitute the values
½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ 
) = 9 109 (1.6 10-19) ²( 
)
2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( 
)
2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷
r₂² -1 = (4.443 10⁸)⁻¹
r2 = 
r2 = 1 m
therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m
