The average velocity of the following 4 velocity measurements will be d) 8.7 m/s
average of given velocities = sum of all velocities divided by number of velocity mentioned in the question
average velocity = ( v1 + v2 + v3 + v4 ) / 4
= ( 9.6 + 8.8 + 7.6 + 8.7 ) / 4 = 8.675 ≈ 8.7 m/s
correct answer d)
The average velocity of the following 4 velocity measurements will be d) 8.7 m/s
To learn more about velocity here :
brainly.com/question/18084516
#SPJ1
Answer:
C ) 1.53
Explanation:
The critical angle of a material is given by the formula
where
c is the critical angle
n is the refractive index
This formula is valid if the second medium is air (which is the case of the problem).
In this problem, we know the critical angle:
Therefore we can rearrange the equation to find the refractive index:
Answer:
(a). The path length is 3.09 m at 30°.
(b). The path length is 188.4 m at 30 rad.
(c). The path length is 1111.5 m at 30 rev.
Explanation:
Given that,
Radius = 5.9 m
(a). Angle 
We need to calculate the angle in radian
We need to calculate the path length
Using formula of path length
(b). Angle = 30 rad
We need to calculate the path length
(c). Angle = 30 rev
We need to calculate the angle in rad
We need to calculate the path length
Hence, (a). The path length is 3.09 m at 30°.
(b). The path length is 188.4 m at 30 rad.
(c). The path length is 1111.5 m at 30 rev.
Explanation:
We Know That
POTENTIAL ENERGY= MASS*g*HEIGHT
When the objects are lifted to same height then the object with heavier mass would have the highest potential energy
.
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.
