Answer:
The wavelength of this wave is 1.01 meters.
Explanation:
The variation in the pressure of helium gas, measured from its equilibrium value, is given by :
..............(1)
The general equation is given by :
...........(2)
On comparing equation (1) and (2) :

Since, 


So, the wavelength of this wave is 1.01 meters. Hence, this is the required solution.
Answer:
Ionic bonds form when a nonmetal and a metal exchange electrons, while covalent bonds form when electrons are shared between two nonmetals. An ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions.
Explanation:
hope this helps!
Answer:
a) 37.8 W
b) 2 Nm
Explanation:
180 g = 0.18 kg
We can also convert 180 revolution per minute to standard angular velocity unit knowing that each revolution is 2π and 1 minute equals to 60 seconds
180 rpm = 180*2π/60 = 18.85 rad/s
We can use the heat specific equation to find the rate of heat exchange of the steel drill and block:

Since the entire mechanical work is used up in producing heat, we can conclude that the rate of work is also 37.8 J/s, or 37.8 W
The torque T required to drill can be calculated using the work equation



Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V
Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:
h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.
In order to calculate the cutoff wavelength we have to consider that Ek=0
in this case h*ν=W
(h*c)/λ=4.52 eV
λ= (h*c)/4.52 eV
λ= (1240 eV*nm)/(4.52 eV)=274.34 nm
From this h*ν = Ek+W; we can calculate the kinetic energy for a radiation wavelength of 198 nm
then we have
(h*c)/(λ)-W= Ek
Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV
Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this acts to slow down the ejected electrons from the catode.