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3 years ago

11

If 20 beats are produce with one second which of the following frequency could possibly be held by two sound waves traveling thr

ough a medium a long a common path at the same time

1 answer:

musickatia [10]3 years ago

8 0

ANY two frequencies whose difference is 20 Hz.

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on a recent adventure trip, Anita Break went rock climbing. Anita was able to perform 1370 J of work in 100 seconds. Determine A

Zanzabum

      Power = (energy) / (time)

   =    (1370 joules) / (100 seconds)

   = 13.7 joules/second

   = 13.7 watts .

That's not an awful lot of power, especially for a strenuous activity like
rock-climbing. Shoot ! Even I could probably perform at that level.

Compare 13.7 watts to the light power coming out of a 20-watt night light.

   13.7 watts = 0.018 horsepower. (rounded)

5 0

2 years ago

(a) An elevator of mass m moving upward has two forces acting on it: the upward force of tension in the cable and the downward f

Katen [24]

Answer: T is greater

Explanation:

Since the elevator is moving against gravity more work will be done on the rope

T= m(g+a)

8 0

2 years ago

Read 2 more answers

A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra

kkurt [141]

Answer:

$V_{ft}= 317 cm/s$

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

$P_i = P_f$

Where:

$P_i=M_cV_{ic} + M_tV_{it}$

$P_f = M_cV_{fc} + M_tV_{ft}$

Now:

$M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}$

Where $M_c$ is the mass of the car, $V_{ic}$ is the initial velocity of the car, $M_t$ is the mass of train, $V_{fc}$ is the final velocity of the car and $V_{ft}$ is the final velocity of the train.

Replacing data:

$(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}$

Solving for $V_{ft}$ :

$V_{ft}= 3.17 m/s$

Changed to cm/s, we get:

$V_{ft}= 3.17*100 = 317 cm/s$

b) The kinetic energy K is calculated as:

K = $\frac{1}{2}MV^2$

where M is the mass and V is the velocity.

So, the initial K is:

$K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2$

$K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2$

$K_i = 22.06 J$

And the final K is:

$K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = 19.61 J$

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0

3 years ago

At which position would a person on earth see the entire lighted half of the moon?

Gemiola [76]

The best position for the person would be outside, under a clear sky, standing up. He should do it sometime between sunset and sunrise, from a day before until a day after the moment of Full Moon.

3 0

2 years ago

Humans living in highly populated areas are more inclined to

Eduardwww [97]

Answer:

they're more inclined to be violent so A

4 0

3 years ago

Read 2 more answers