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GenaCL600 [577]
3 years ago
15

The pitch made by a musical instrument is dependent on the frequency of the wave. Frequency is the inverse of period. What impac

t do you think the frequency of a wave has on the wave speed?
Physics
2 answers:
Blizzard [7]3 years ago
4 0

I think the frequency of the wave has <em>no effect</em> whatsoever on the wave speed.  The speed of a wave completely depends on the characteristics of the medium it's traveling through.  

Physical properties of the medium determine the speed of mechanical waves, like sound, ocean waves, seismic waves, Slinky waves etc.  If it depended on the frequency of the waves, then the sound of the band or orchestra would be all smeared out when it reaches you ... you'd hear the drums and bass first, and the flutes and trumpets last.

Electrical properties of the medium determine the speed of electromagnetic waves, like radio, light, and X-rays. If it depended on the frequency of the waves, then you couldn't watch a football game in the stadium or on TV.  The reds, oranges, and yellows would reach your eyes first, and then the greens, blues, and violets would arrive later.  The action would be all smeared out, and things that happened first might reach  your eyes last.  What a mess !

polet [3.4K]3 years ago
4 0

Answer:

it has no affect at all .. doesnt change

Explanation:

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A 50.0 N box sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at on
Vikki [24]

Answer:

-4.0 N

Explanation:

Since the force of friction is the only force acting on the box, according to Newton's second law its magnitude must be equal to the product between mass (m) and acceleration (a):

F_f = ma (1)

We can find the mass of the box from its weight: in fact, since the weight is W = 50.0 N, its mass will be

m=\frac{W}{g}=\frac{50.0 N}{9.8 m/s^2}=5.1 kg

And we can fidn the acceleration by using the formula:

a=\frac{v-u}{t}

where

v = 0 is the final velocity

u = 1.75 m/s is the initial velocity

t = 2.25 s is the time the box needs to stop

Substituting, we find

a=\frac{0-1.75 m/s}{2.25 s}=-0.78 m/s^2

(the acceleration is negative since it is opposite to the motion, so it is a deceleration)

Therefore, substituting into eq.(1) we find the force of friction:

F_f = (5.1 kg)(-0.78 m/s^2)=-4.0 N

Where the negative sign means the direction of the force is opposite to the motion of the box.

6 0
3 years ago
A frictionless pendulum is made with a bob of mass 19.7 kg. The bob is held at height = 0.934 meter above the bottom of its traj
Alika [10]

Answer:

265 J

Explanation:

Energy=PE+KE=mgh+ 0.5mv^{2} where KE is kinetic energy, PE is potential energy, m is the mass of an object, v is the speed, h is the height and g is acceleration due to gravity.

Substituting 19.7 Kg for mass, 0.934 for h, 2.93 for v and 9.81 for g then

Energy=19.7(9.81*0.934+0.5*2.93^{2})=265.063303\approx 265 J

4 0
3 years ago
An object of mass kg is released from rest m above the ground and allowed to fall under the influence of gravity. Assuming the f
IgorLugansk [536]

Answer:

Explanation:

From, the given information: we are not given any value for the mass, the proportionality constant and the distance

Assuming that:

the mass = 5 kg and the proportionality constant = 50 kg

the distance of the mass above the ground x(t) = 1000 m

Let's recall that:

v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}

Similarly, The equation of mption:

x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})

replacing our assumed values:

where v_=0 \ and \ g= 9.81

x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})

x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

So, when the object hits the ground when x(t) = 1000

Then from above derived equation:

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

1000= 0.981t-0.981(1-e^{-(10)t}) \ m

By diregarding e^{-(10)t} \ m

1000= 0.981t-0.981

1000 + 0.981 = 0.981 t

1000.981 = 0.981 t

t = 1000.981/0.981

t = 1020.36 sec

7 0
3 years ago
If the force of a golf club on a golf ball is 200 N forward, what will the force of the ball on the club be? A. 200 N forward B.
Korolek [52]

The ball should put 200 N of force towards the golfer.

Newton's Third Law is every action has an equal and opposite reaction.

It's the ball exerting 200 N of force towards the club as well, but the opposite reaction is that it flies away.

8 0
4 years ago
Read 2 more answers
As you finish listening to your favorite compact disc (CD), the CD in the player slows down to a stop. Assume that the CD spins
Colt1911 [192]

Answer:

A. w=8.3rev/s

a_v=3.21rev/s^2

B. rev=10.79rev

Explanation:

A.

500rpm

500rpm*\frac{1minute}{60seg}=8.33\frac{rev}{s}

the time is about 2.60s knowing the acceleration is:

a_v=\frac{w}{t}

so:

w=8.3rev/s

a_v=\frac{8.33 rev/s}{2.60s}=3.21rev/s^2

B.

The CD stop so the final velocity is vf=0 so:

Total revolutions:

8.3rev/s*2.60s=21.58rev

21.58rev/2=10.79 rev

3 0
4 years ago
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