Test:
Performing a Litmus Test
Result:
Litmus paper gives the user a general indication of acidity or alkalinity as it correlates to the shade of red or blue that the paper turns.
- To test the pH of a substance, dip a strip of litmus paper into the solution or use a dropper or pipette to drip a small amount of solution onto the litmus paper.
- Blue litmus paper can indicate an acid with a pH between 4 and 5 or lower.
- Red litmus paper can show a base with a pH greater than 8.
- If a solution has a pH between 5 and 8, it will show little color change on the litmus paper.
- A base tested with blue litmus paper will not show any color change, nor will an acid tested with red litmus paper register a change in color.
Answer:
λ = 482.05 nm
Explanation:
The diffraction phenomenon and the diffraction grating is described by the expression
d sin θ = m λ
where d is the distance between two consecutive slits, λ the wavelength and m an integer representing the order of diffraction
in this case they indicate the distance between slits, the angle and the order of diffraction
λ = 
d sin θ / m
let's calculate
λ = 1.00 10⁻⁶ sin 74.6 / 2
λ = 4.82048 10⁻⁷ m
Let's reduce to nm
λ = 4.82048 10⁻⁷ m (10⁹ nm / 1 m)
λ = 482.05 nm
Answer:
Vb = k Q / r r <R
Vb = k q / R³ (R² - r²) r >R
Explanation:
The electic potential is defined by
ΔV = - ∫ E .ds
We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product
VB - VA = - ∫ E dr
Let's substitute every equation they give us and we find out
r> R
Va = - ∫ (k Q / r²) dr
-Va = - k Q (- 1 / r)
We evaluate with it Va = 0 for r = infinity
Vb = k Q / r r <R
We perform the calculation of the power with the expression of the electric field that they give us
Vb = - int (kQ / R3 r) dr
We integrate and evaluate from the starting point r = R to the final point r <R
Vb = ∫kq / R³ r dr
Vb = k q / R³ (R² - r²)
This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity
Answer:
center of mass of the two masses will lie at x = 2.52 cm
center of gravity of the two masses will lie at x = 2.52 cm
So center of mass is same as center of gravity because value of gravity is constant here
Explanation:
Position of centre of mass is given as
here we have
now we have
so center of mass of the two masses will lie at x = 2.52 cm
now for center of gravity we can use
here we have
now we have
So center of mass is same as center of gravity because value of gravity is constant here
