Answer:
<em>d. unchanged.</em>
Explanation:
The frequency of a wave is dependent on the speed of the wave and the wavelength of the wave. The frequency is characteristic for a wave, and does not change with distance. This is unlike the amplitude which determines the intensity, which decreases with distance.
In a wave, the velocity of propagation of a wave is the product of its wavelength and its frequency. The speed of sound does not change with distance, except when entering from one medium to another, and we can see from
v = fλ
that the frequency is tied to the wave, and does not change throughout the waveform.
where v is the speed of the sound wave
f is the frequency
λ is the wavelength of the sound wave.
Answer:
16 cm
Explanation:
For protons:
Energy, E = 300 keV
radius of orbit, r1 = 16 cm
the relation for the energy and velocity is given by

So,
.... (1)
Now,

Substitute the value of v from equation (1), we get

Let the radius of the alpha particle is r2.
For proton
So,
... (2)
Where, m1 is the mass of proton, q1 is the charge of proton
For alpha particle
So,
... (3)
Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle
Divide equation (2) by equation (3), we get

q1 = q
q2 = 2q
m1 = m
m2 = 4m
By substituting the values

So, r2 = r1 = 16 cm
Thus, the radius of the alpha particle is 16 cm.
Answer:
Explanation:
The only thing I can figure you need here is the accleration of the sled. The equation we need to find this is Newton's Second Law that says that sum of the forces acting on an object is equal to the object's mass times its acceleration. For us, that looks like this because of the friction working against the sled:
F - f = ma but of course it's much more involved than that simple equation! We have the F value as 230 N, and we have the mass as 105, but we do not have the frictional force, f, and we need it to solve for a in the above equation. We know that
f = μ
where μ is the coefficient of friction, and
is the normal force, aka weight of the object. We will use the coefficient of friction and find the weight in order to fill in for f:
so
so the weight of the sled is
1.0 × 10³ with the correct number of sig dig there. Now to find f:
f = (.025)(1.0 × 10³) so
f = 25 to the correct number of sig fig. Now on to our "real" equation:
F - f = ma and
230 - 25 = 105a. We have to do the subtraction first, round, and then divide since the rules for addition and subtraction are different from the rules for dividing and multiplying.
230 - 25 will round to the tens place giving us 210. Then
210 = 105a. 210 has 2 sig figs in it while 105 has 3, so we will divide and round to 2 sig fig:
a = 2.0 m/sec²
Answer:
longitudinal wave
Explanation:
it is perpendicular to the direction of the wave
an electric current is a flow of electric charge in electric circuits this is carried by moving electrons in a wire and an electric circuit is an electrical network of electrical components and model of interconnection consisting electrical elements