an object is placed at 19.3 cm in front of a diverging lens with a focal length of -18.7 cm. what is the magnification? enter a
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Answer:
x(t)=0.337sin((5.929t)
Explanation:
A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2 m/sec, find the position of the mass after t seconds.
Solution. Let x(t) denote the position of the mass at time t. Then x satisfies the differential equation
Definition of parameters
m=mass 3kg
k=force constant
e=extension ,m
ω =angular frequency
k=90/1.6=56.25N/m
ω^2=k/m= 56.25/1.6
ω^2=35.15625
ω=5.929
General solution will be
differentiating x(t)
dx(t)=-5.929c1sin(5.929t)+5.929c2cos(5.929t)
when x(0)=0, gives c1=0
dx(t0)=2m/s gives c2=0.337
Therefore, the position of the mass after t seconds is
x(t)=0.337sin((5.929t)
The total distance covered by the sound wave is twice the distance between the camera and the subject (because the wave has to reach the subject and then travel back to the camera), so 2L, where L=3.42 m. The speed of sound is v=343 m/s. It is a uniform linear motion, so we can use the basic relationship between space (S), time (t) and velocity (v) to find the time the wave needs to return to the camera:
Answer: The required distance is given by
Explanation: The sound intensity in dB is given by the formula
where is the hearing threshold in absolute units and
is the absolute intensity of the sound which depends on the distance. In general, for two distances
and
we have that
Now let us take and let
be the required distance. We have
Exponentiating these equations we obtain
Dividing them
Using the previously stated identity
Now if we use the given example where we have