Answer:
The velocity of the photo electron is
.
Explanation:
Given that,
Supplied energy, 
Minimum energy of the electron to escape from the metal, 
We need to find the velocity of the photo electron. The energy supplied by the photon is equal to the sum of minimum escape energy and the kinetic energy of the escaping electron. So,

The formula of kinetic energy is given by :

So, the velocity of the photo electron is
.
Answer:
Magnetic force, 
Explanation:
Given that,
A beryllium-9 ion has a positive charge that is double the charge of a proton, 
Speed of the ion in the magnetic field, 
Its velocity makes an angle of 61° with the direction of the magnetic field at the ion's location.
The magnitude of the field is 0.220 T.
We need to find the magnitude of the magnetic force on the ion. It is given by :

So, the magnitude of magnetic force on the ion is
.
Base in your question about the magnetic field of the Earth near the equator where as its almost horizontally to the north and has magnitude of B=0.5x10^-4t, the answer is <span>Velocity of electron will be westwards.</span>
Answer:
x ’= 368.61 m, y ’= 258.11 m
Explanation:
To solve this problem we must find the projections of the point on the new vectors of the rotated system θ = 35º
x’= R cos 35
y’= R sin 35
The modulus vector can be found using the Pythagorean theorem
R² = x² + y²
R = 450 m
we calculate
x ’= 450 cos 35
x ’= 368.61 m
y ’= 450 sin 35
y ’= 258.11 m
Answer:
306 m/s
Explanation:
Law of conservation of momentum
m1v1 + m2v2 = (m1+m2)vf
m1 is the bullet's mass so it is 0.1 kg
v1 is what we're trying to solve
m2 is the target's mass so it is 5.0 kg
v2 is the targets velocity, and since it was stationary, its velocity is zero
vf is the velocity after the target is struck by the bullet, so it is 6.0 m/s
plugging in, we get
(0.1 kg)(v1) + (5.0 kg)(0 m/s) = (0.1 kg + 5.0 kg)(6.0 m/s)
(0.1)(v1) + 0 = 30.6
(0.1)(v1) = 30.6
v1 = 306 m/s