Answer:
The speed after being pulled is 2.4123m/s
Explanation:
The work realize by the tension and the friction is equal to the change in the kinetic energy, so:
(1)
Where:

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.
Additionally, the kinetic energy is equal to
, so if the initial velocity
is equal to zero, the initial kinetic energy
is equal to zero.
Then, replacing the values on the equation and solving for
, we get:


So, the speed after being pulled 3.2m is 2.4123 m/s
Explanation:
Given that,
Wavelength of the light, 
Work function of sodium, 
The kinetic energy of the ejected electron in terms of work function is given by :

The formula of kinetic energy is given by :

Hence, this is the required solution.
Answer:
Perpendicular to the electric field and magnetic field
Explanation:
Electromagnetic waves are transverse waves composed by the perpendicular oscillating electric and magnetic fields.
EM waves have both Electrical and magnetic features.
they travel in the velocity of light (3*10⁸ ms⁻¹)
they does not require any media to travel. It has two perpendicular electric field and the magnetic field which are perpendicular to each other
They travel perpendicular to each of those electric and magnetic fields.
Answer:
d = 10.076 m
Explanation:
We need to obtain the velocity of the ball in the y direction
Vy = 24.5m/s * sin(35) = 14.053 m/s
To obtain the distance, we use the formula
vf^2 = v0^2 -2*g*d
but vf = 0
d = -vo^2/2g
d = (14.053)^2/2*(9.8) = 10.076 m