Answer:
0.7515875 eV

Explanation:
f = Maximum frequency = 
h = Planck's constant = 
W = Work function = 2.52 eV
Converting to Joules

Maximum photon energy is given by

Maximum Kinetic energy is given by

Converting to eV

The maximum kinetic energy of electrons ejected from this surface is 0.7515875 eV

The range of frequencies for which no electrons are ejected is

The basic structure would be:
Reactants → Products
Answer:
a) 
b) 
c) 
d)
or 18.3 cm
Explanation:
For this case we have the following system with the forces on the figure attached.
We know that the spring compresses a total distance of x=0.10 m
Part a
The gravitational force is defined as mg so on this case the work donde by the gravity is:

Part b
For this case first we can convert the spring constant to N/m like this:

And the work donde by the spring on this case is given by:

Part c
We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

And if we solve for the initial velocity we got:

Part d
Let d1 represent the new maximum distance, in order to find it we know that :

And replacing we got:

And we can put the terms like this:

If we multiply all the equation by 2 we got:

Now we can replace the values and we got:


And solving the quadratic equation we got that the solution for
or 18.3 cm because the negative solution not make sense.
Answer:
3.82 Ns
Explanation:
Time varying horizontal Force is given as
F(t) = A t⁴ + B t²
F(t) = 4.50 t⁴ + 8.75 t²
Impulse imparted is given as





(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.
(b) The maximum height above the ground reached by the ball is 8.6 m.
(c) The distance off course the ball would be carried is 0.38 m.
(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
<h3>
Horizontal and vertical components of the ball's velocity</h3>
Vx = Vcosθ
Vx = 39.7 x cos(17.8)
Vx = 37.8 m/s
Vy = Vsin(θ)
Vy = 39.7 x sin(17.8)
Vy = 12.14 m/s
<h3>Maximum height reached by the ball</h3>

Maximum height above ground = 7.51 + 1.09 = 8.6 m
<h3>Distance off course after 2 second </h3>
Upward speed of the ball after 2 seconds, V = V₀y - gt
Vy = 12.14 - (2x 9.8)
Vy = - 7.46 m/s
Horizontal velocity will be constant = 37.8 m/s
Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

<h3>Resultant speed of the ball and crosswind</h3>

<h3>Distance off course the ball would be carried</h3>
d = Δvt = (38.72 - 38.53) x 2
d = 0.38 m
The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
Learn more about projectiles here: brainly.com/question/11049671