Answer:
given
y=6.0sin(0.020px + 4.0pt)
the general wave equation moving in the positive directionis
y(x,t) = ymsin(kx -?t)
a) the amplitude is
ym = 6.0cm
b)
we have the angular wave number as
k = 2p /?
or
? = 2p / 0.020p
=1.0*102cm
c)
the frequency is
f = ?/2p
= 4p/2p
= 2.0 Hz
d)
the wave speed is
v = f?
= (100cm)(2.0Hz)
= 2.0*102cm/s
e)
since the trignometric function is (kx -?t) , sothe wave propagates in th -x direction
f)
the maximum transverse speed is
umax =2pfym
= 2p(2.0Hz)(6.0cm)
= 75cm/s
g)
we have
y(3.5cm ,0.26s) = 6.0cmsin[0.020p(3.5) +4.0p(0.26)]
= -2.0cm
The work W done by the electric field in moving the proton is equal to the difference in electric potential energy of the proton between its initial location and its final location, therefore:
where q is the charge of the proton,
, with
being the elementary charge, and
and
are the initial and final voltage.
Substituting, we get (in electronvolts):
and in Joule:
Answer:
The first is the electric field, which describes the force acting on a stationary charge and gives the component of the force that is independent of motion. The magnetic field, in contrast, describes the component of the force that is proportional to both the speed and direction of charged particles.
Answer:
F = 351×10³lb
Explanation:
Given the density
ρg = 64.6lb/ft³
Diameter d = 12ft
The tank is horizontally cylindrical. The vertical distance from the top to the bottom of the tank is h = 12ft
The pressure in the tank is
P = ρgh = 64.6 × 12 = 775.2lb/ft²
The force exerted on one end of the tank is therefore F = PA = 775.2 × πd² = 775.2π×12²
F = 351×10³lb.
