Answer:
= 5/9
Explanation:
This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.
B = ρ_liquid g V_liquid
let's write the translational equilibrium condition
B - W = 0
let's use the definition of density
ρ_body = m / V_body
m = ρ_body V_body
W = ρ_body V_body g
we substitute
ρ_liquid g V_liquid = ρ_body g V_body
In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar
V = A h_bogy
Thus
we substitute
5/9 = 
Use v = u + at
Message me if you need more help
Answer:
15 cm
Explanation:
= Diameter of the coin = 15 mm
= Diameter of the image of coin = 5 mm
= distance of the coin from mirror = 15 cm
= distance of the image of coin from mirror = ?
Using the equation


= - 5 cm
= radius of curvature
Using the mirror equation


= - 15 cm
Answer:
true
Explication:
The acceleration of an object depends on the mass of the object, and the amount of force applied
Explanation:
Given that,
The mean kinetic energy of the emitted electron, 
(a) The relation between the kinetic energy and the De Broglie wavelength is given by :



(b) According to Bragg's law,

n = 1
For nickel, 



As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.