Answer:

Explanation:
For the simple pendulum problem we need to remember that:
,
where
is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:
,
where
is the angular frequency.
There is also an equation that relates the oscillation period and the angular frequeny:
,
where T is the oscillation period. Now, we can easily solve for L:

Answer:
<h2>3.36J</h2>
Explanation:
Step one:
given data
mass m= 1.3kg
distance moved s= 2.8m
opposing frictional force= 0.34N
assume g= 9.81m/s^2
we know that work done= force *distance moved
1. work done to push the book= 1.55*2.8=4.34J
2. Work against friction = force of friction x distance
= 0.34*2.8=0.952J
Step two:
the work done on the book is the net work, which is
Network done= work done to push the book- Work against friction
Network done= 4.32-0.952=3.36J
<u>Therefore the work of the 1.55N 3.36J</u>
Answer:
75.36 mph
Explanation:
The distance between the other car and the intersection is,
The distance between the police car and the intersection is,
(Negative sign indicates that he is moving towards the intersection)
Therefore the distance between them is given by,
The rate of change is,
Now finding
when
from (1) we have
The officer's radar gun indicates 25 mph pointed at the other car then,
when
from
From (2) we get
Hence the speed of the car is 
Answer:
The shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm
Explanation:
Given;
wavelength of ultraviolet light, λ = 270 nm
work function of the metal, φ = 2.3 eV = 2.3 x 1.602 x 10⁻¹⁹ J = 3.685 x 10⁻¹⁹ J
The energy of the ultraviolet light is given by;

The energy of the incident light is related to kinetic energy of the electron and work function of the metal by the following equation;
E = φ + K.E
K.E = E - φ
K.E = (7.362 x 10⁻¹⁹ J) - (3.685 x 10⁻¹⁹ J )
K.E = 3.677 x 10⁻¹⁹ J
K.E = ¹/₂mv²
mv² = 2K.E
velocity of the electron is given by;

the shortest de Broglie wavelength for the electrons is given by;

Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm