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
Answer:
5.5 × 10^14 Hz or s^-1
no orange light has less frequency so no photoelectric effect
Explanation:
hf = hf0 + K.E
HERE h is Planck 's constant having value 6.63 × 10 ^-34 J s
f is frequency of incident photon and f0 is threshold frequency
hf0 = hf- k.E
6.63 × 10 ^-34 × f0 = 6.63 × 10 ^-34× 6.66 × 10^14 - 7.74× 10^-20
6.63 × 10 ^-34 × f0 = 3.64158×10^-19
f0 = 3.64158×10^-19/ 6.63 × 10 ^-34
f0 = 5.4925 × 10^14
f0 =5.5 × 10^14 Hz or s^-1
frequency of orange light is 4.82 × 10^14 Hz which is less than threshold frequency hence photo electric effect will not be observed for orange light
Answer:
Explanation:
Let v be the linear velocity , ω be the angular velocity and I be the moment of inertia of the the puck.
Kinetic energy ( linear ) = 1/2 mv²
Rotational kinetic energy = 1/2 I ω²
I = 1/2 m r² ( m and r be the mass and radius of the puck )
Rotational kinetic energy = 1/2 x1/2 m r² ω²
= 1/4 m v² ( v = r ω )
Total energy
= Kinetic energy ( linear ) + Rotational kinetic energy
= 1/2 mv² + 1/4 m v²
= 3/4 mv²
rotational K E / Total K E = 1/4 m v² / 3/4 mv²
= 1 /3
So 1 /3 rd of total energy is rotational K E.
Answer:
An Element
Explanation:
Such substance cannot be an element because an element cannot be chemically disintegrated (i.e it cannot be disintegrated via chemical reaction).
Answer:
(A) 374.4 J
(B) -332.8 J
(C) 0 J
(D) 41.6 J
(E) 351.8 J
Explanation:
weight of carton (w) = 128 N
angle of inclination (θ) = 30 degrees
force (f) = 72 N
distance (s) = 5.2 m
(A) calculate the work done by the rope
- work done = force x distance x cos θ
- since the rope is parallel to the ramp the angle between the rope and
the ramp θ will be 0
work done = 72 x 5.2 x cos 0
work done by the rope = 374.4 J
(B) calculate the work done by gravity
- the work done by gravity = weight of carton x distance x cos θ
- The weight of the carton = force exerted by the mass of the carton = m x g
- the angle between the force exerted by the weight of the carton and the ramp is 120 degrees.
work done by gravity = 128 x 5.2 x cos 120
work done by gravity = -332.8 J
(C) find the work done by the normal force acting on the ramp
- work done by the normal force = force x distance x cos θ
- the angle between the normal force and the ramp is 90 degrees
work done by the normal force = Fn x distance x cos θ
work done by the normal force = Fn x 5.2 x cos 90
work done by the normal force = Fn x 5.2 x 0
work done by the normal force = 0 J
(D) what is the net work done ?
- The net work done is the addition of the work done by the rope, gravitational force and the normal force
net work done = 374.4 - 332.8 + 0 = 41.6 J
(E) what is the work done by the rope when it is inclined at 50 degrees to the horizontal
- work done by the rope= force x distance x cos θ
- the angle of inclination will be 50 - 30 = 20 degrees, this is because the ramp is inclined at 30 degrees to the horizontal and the rope is inclined at 50 degrees to the horizontal and it is the angle of inclination of the rope with respect to the ramp we require to get the work done by the rope in pulling the carton on the ramp
work done = 72 x 5.2 x cos 20
work done = 351.8 J
