For a wave:
v = fλ
v is the velocity, f is the frequency, and λ is the wavelength.
Assuming the velocity of the wave doesn't change...
If you increase its frequency, its wavelength will shorten.
Answer:
linear charge density = -9.495 ×
C/m
Explanation:
given data
revolutions per second = 1.80 ×
radius = 1.20 cm
solution
we know that when proton to revolve around charge wire then centripetal force is require to be in orbit of radius around provide by electric force
so
- q × E = m × w² × r ..................1
- 9 ×
×
q = m × w² × r ............2
and w =
w =
w = 1.80 ×
×
w = 11304000 rad/s
so here from equation 2
- 9 ×
×
1.80 ×
= 1.672 ×
× 11304000² × 0.0120
linear charge density = -9.495 ×
C/m
Answer:
The work done on the canister by the 5.0 N force during this time is
54.06 Joules.
Explanation:
Let the initial kinetic energy of the canister be
KE₁ =
=
= 19.44 J in the x direction
Let the the final kinetic energy of the canister be
KE₂ =
=
= 73.5 J in the y direction
Therefore from the Newton's first law of motion, the effect of the force is the change of momentum and the difference in energy between the initial and the final
= 73.5 J - 19.44 J = 54.06 J
4. The Coyote has an initial position vector of
.
4a. The Coyote has an initial velocity vector of
. His position at time
is given by the vector

where
is the Coyote's acceleration vector at time
. He experiences acceleration only in the downward direction because of gravity, and in particular
where
. Splitting up the position vector into components, we have
with


The Coyote hits the ground when
:

4b. Here we evaluate
at the time found in (4a).

5. The shell has initial position vector
, and we're told that after some time the bullet (now separated from the shell) has a position of
.
5a. The vertical component of the shell's position vector is

We find the shell hits the ground at

5b. The horizontal component of the bullet's position vector is

where
is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for
:
