Answer:
d. a=−k/mx
Explanation:
To know what is the correct expression for the acceleration you take into account the second Newton law, that is:
( 1 )
next, you equal the expression ( 1 ) to the force in a mass-string system, that is F=-kx.
hence, the acceleration is:
d. a=−kmx
Answer:
a) 1.2*10^-7 m
b) 1.0*10^-7 m
c) 9.7*10^-8 m
d) ultraviolet region
Explanation:
To find the different wavelengths you use the following formula:
RH: Rydberg constant = 1.097 x 10^7 m^−1.
(a) n=2
(b)
(c)
(d) The three lines belong to the ultraviolet region.
Answer:
Proton: v=0.689 m/s
Neutron: v=0.688 m/s
Electron: v=1265.078 m/s
Alpha particle: v=0.173 m/s
Explanation:
De Broglie equation allows you to calculate the “wavelength” of an electron or any other particle or object of mass m that moves with velocity v:
λ=
h is the Planck constant: 6.626×10⁻³⁴
We know that the wavelength of the particle is 575 nm (575×10⁻⁹m), so we find the velocity v for each particle:
λ=
v=h÷(mλ)
<u>Proton:</u>
m=1.673×10⁻²⁴ g · =1.673×10⁻²⁷ kg
v=h÷(mλ)
v=6.626×10⁻³⁴÷(1.673×10⁻²⁷ kg×575×10⁻⁹m)
v=0.689 m/s
<u>Neutron:</u>
m=1.675×10⁻²⁴ g · =1.675×10⁻²⁷ kg
v=h÷(mλ)
v=6.626×10⁻³⁴÷(1.675×10⁻²⁷ kg×575×10⁻⁹m)
v=0.688 m/s
<u>Electron:</u>
m= 9.109×10⁻²⁸ g · =9.109×10⁻³¹ kg
v=h÷(mλ)
v=6.626×10⁻³⁴÷(9.109×10⁻³¹ kg×575×10⁻⁹m)
v=1265.078 m/s
<u>Alpha particle:</u>
m=6.645×10⁻²⁴ g · =6.645×10⁻²⁷ kg
v=h÷(mλ)
v=6.626×10⁻³⁴÷(6.645×10⁻²⁷ kg×575×10⁻⁹m)
v=0.173 m/s
I think this is right I hope this is right for you
