The density of sample is 5 g/cm3
Given:
volume of sample = 20 cm3
mass of sample = 100 grams
To Find:
density of sample
Solution: Density is the measure of how much “stuff” is in a given amount of space. For example, a block of the heavier element lead (Pb) will be denser than the softer, lighter element gold (Au). A block of Styrofoam is less dense than a brick. It is defined as mass per unit volume
density = mass/volume
d = 100/20
d = 5 g/cm3
So, density of sample is 5 g/cm3
Learn more about Density here:
brainly.com/question/1354972
#SPJ4
The goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground
Explanation:
Consider the vertical motion of ball,
We have equation of motion v = u + at
Initial velocity, u = u sin θ
Final velocity, v = 0 m/s
Acceleration = -g
Substituting
v = u + at
0 = u sin θ - g t
![t=\frac{usin\theta }{g}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Busin%5Ctheta%20%7D%7Bg%7D)
This is the time of flight.
Consider the horizontal motion of ball,
Initial velocity, u = u cos θ
Acceleration, a =0 m/s²
Time,
Substituting
s = ut + 0.5 at²
![s=ucos\theta \times \frac{usin\theta }{g}+0.5\times 0\times (\frac{usin\theta }{g})^2\\\\s=\frac{u^2sin\theta cos\theta}{g}\\\\s=\frac{u^2sin2\theta}{2g}](https://tex.z-dn.net/?f=s%3Ducos%5Ctheta%20%5Ctimes%20%5Cfrac%7Busin%5Ctheta%20%7D%7Bg%7D%2B0.5%5Ctimes%200%5Ctimes%20%28%5Cfrac%7Busin%5Ctheta%20%7D%7Bg%7D%29%5E2%5C%5C%5C%5Cs%3D%5Cfrac%7Bu%5E2sin%5Ctheta%20cos%5Ctheta%7D%7Bg%7D%5C%5C%5C%5Cs%3D%5Cfrac%7Bu%5E2sin2%5Ctheta%7D%7B2g%7D)
This is the range.
In this problem
u = 30 m/s
g = 9.81 m/s²
θ = 45° - For maximum range
Substituting
![s=\frac{30^2\times sin(2\times 45)}{2\times 9.81}=45.87m](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B30%5E2%5Ctimes%20sin%282%5Ctimes%2045%29%7D%7B2%5Ctimes%209.81%7D%3D45.87m)
Maximum horizontal distance traveled by ball without touching ground is 45.87 m, which is less than 95 m.
So the goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground
Answer:
(a) 5.04 eV (B) 248.14 nm (c) ![1.21\times 10^{15}Hz](https://tex.z-dn.net/?f=1.21%5Ctimes%2010%5E%7B15%7DHz)
Explanation:
We have given Wavelength of the light \lambda = 240 nm
According to plank's rule ,energy of light
![E = h\nu = \frac{hc}{}\lambda](https://tex.z-dn.net/?f=E%20%3D%20h%5Cnu%20%3D%20%5Cfrac%7Bhc%7D%7B%7D%5Clambda)
![E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV](https://tex.z-dn.net/?f=E%20%3D%20h%5Cnu%20%3D%20%5Cfrac%7B6.67%5Ctimes%2010%5E%7B-34%7D%20J.s%5Ctimes%203%5Ctimes%2010%5E%7B8%7Dm%2Fs%7D%7B%20240%5Ctimes%2010%5E%7B-9%7D%20m%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7DJ%2FeV%7D%3D%205.21%20eV)
Maximum KE of emitted electron i= 0.17 eV
Part( A) Using Einstien's equation
, here
is work function.
= 5.21 eV-0.17 eV = 5.04 eV
Part( B) We have to find cutoff wavelength
![\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}](https://tex.z-dn.net/?f=%5CPhi%20_%7B0%7D%20%3D%20%5Cfrac%7Bhc%7D%7B%5Clambda_%7Bcuttoff%7D%7D)
![\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }](https://tex.z-dn.net/?f=%5Clambda_%7Bcuttoff%7D%3D%20%5Cfrac%7Bhc%7D%7B%5CPhi%20_%7B0%7D%20%7D)
![\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm](https://tex.z-dn.net/?f=%5Clambda_%7Bcuttoff%7D%3D%20%5Cfrac%7B6.67%5Ctimes%2010%5E%7B-34%7D%20J.s%5Ctimes%203%5Ctimes%2010%5E%7B8%7Dm%2Fs%7D%7B5.04%20eV%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7DJ%2FeV%20%7D%3D248.14%20nm)
Part (C) In this part we have to find the cutoff frequency
![\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz](https://tex.z-dn.net/?f=%5Cnu%20%3D%20%5Cfrac%7Bc%7D%7B%5Clambda_%7Bcuttoff%7D%7D%3D%20%5Cfrac%7B3%5Ctimes%2010%5E%7B8%7Dm%2Fs%7D%7B248.14%20%5Ctimes%2010%5E%7B-19%7D%20m%20%7D%3D%201.21%5Ctimes%2010%5E%7B15%7D%20Hz)
M=2.45 because you multiply out the equation on the right and divide by 10