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2 years ago

When the amplitudes of waves are equal, which frequency waves have the most energy?

Physics

1 answer:

Molodets [167]2 years ago

5 0

Answer:

higher frequency waves will have more energy

Explanation:

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The sound from a single source can reach point O by two different paths. One path is 20.0 m long and the second path is 21.0 m l

aleksandrvk [35]

Answer:

minimum frequency = 170 Hz

Explanation:

given data

One path long = 20 m

second path long = 21 m

speed of sound = 340 m/s

solution

we get here destructive phase that is path difference of minimum $\frac{\lambda}{2}$

here λ is the wavelength of the wave

so path difference will be

21 - 20 = $\frac{\lambda}{2}$

λ = 2 m

and

velocity that is express as

velocity = frequency × wavelength .............1

frequency = $\frac{340}{2}$

minimum frequency = 170 Hz

7 0

3 years ago

At what speed do a bicycle and its rider, with a combined mass of 100 kg , have the same momentum as a 1600 kg car traveling at

noname [10]

$Given:\\m_b=100kg\\m_c=1600kg\\v_c=5.2 \frac{m}{s} \\p_b=p_c\\\\Find:\\v_b=?\\\\Solution:\\\\p_b=p_c\\\\p=mv\\\\m_bv_b=m_cv_c\Rightarrow v_b= \frac{m_cv_c}{m_b} \\\\v_b= \frac{1600kg\cdot5.2 \frac{m}{s} }{100kg} =83.2 \frac{m}{s}$

3 0

3 years ago

A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the

lana [24]

(a) Let's convert the final speed of the car in m/s:
$v_f = 61 km/h = 16.9 m/s$
The kinetic energy of the car at t=19 s is
$K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J$

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
$P= \frac{W}{\Delta t}$
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
$P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W$

(c) The instantaneous power is given by
$P_i = Fv_f$
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
$a= \frac{v_f-v_i}{\Delta t}= \frac{16.9 m/s}{19 s}=0.89 m/s^2$
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
$P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W$

5 0

3 years ago

50 POINTS !!!

shepuryov [24]

Answer:

a) I = 464 kg m², b) K = 631 .6 J, c) v = 8.25 m / s

Explanation:

a) the moment of inertia of point particles is

I = ∑ m_i r_i²

in this case

I = 8 5² + 3 (-2) ² + 7 (-6) ²

I = 464 kg m²

b) The kinetic energy is

K = ½ I w²

K = ½ 464 1.65²

K = 631 .6 J

c) linear and angular velocity are related

v = w r

v = 1.65 5

v = 8.25 m / s

8 0

2 years ago

Can a material have negative permittivity?

sleet_krkn [62]

No it can't it's material

3 0

3 years ago

When the amplitudes of waves are equal, which frequency waves have the most energy?​

When the amplitudes of waves are equal, which frequency waves have the most energy?