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3 years ago

7

Any two waves that meet as they travel through the same medium will interact. If the two waves are in phase their crests will co

incide and they will experience _____ interference, producing a wave with a(n) _____ amplitude.

1 answer:

MrRissso [65]3 years ago

6 0

Answer:

constructive

alpitude

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What observations tell you that a car is accelerating

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When the car moves and makes a sound that is louder that when the car is just sitting there

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3 years ago

uphill at a rate of 2.5 mi/h from the base of a 6-mi trail. At the same time, Edwin walks downhill at a rate of 3.5 mi/h from th

uranmaximum [27]

<h2>After 1 hour they meet.</h2>

Explanation:

Distance between them = 6 miles

Speed of uphill person = 2.5 miles per hour

Speed of downhill person = 3.5 miles per hour

Relative velocity = 2.5 - ( -3.5 ) = 6 miles per hour

We know

Displacement = Velocity x Time

6 = 6 x Time taken

Time taken = 1 hour

After 1 hour they meet.

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4 years ago

Why is there so much variation in human skin coloration? a. This occurred because humans underwent natural selection, which resu

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Answer:

<u>e. All of these.</u>

Explanation:

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3 years ago

A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an

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Answer:

a) $T=0.0031416s$

b) $v_{max}=6.283\frac{m}{s}$

c) $E=0.1974J$

d) $F=80N$

e) $F=40N$

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by $X=Acos(\omega t +\phi)$ (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

$\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)$ (2)

For the acceleration

$\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)$ (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

$A\omega^2=8x10^{3}\frac{m}{s^2}$

Since we know the amplitude A=0.002m we can solve for $\omega$ like this:

$\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}$

And we with this value we can find the period with the following formula

$T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s$

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

$v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}$

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

$E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J$

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens $cos(\omega t +\phi)=1$ , and we also know that the maximum acceleration is given by::

$|\frac{d^2X}{dt^2}|=A\omega^2$

So then we have that:

$F=ma=mA\omega^2$

And since we have everything we can find the force

$F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N$

6) Part e

When the mass it's at the half of it's maximum displacement the term $cos(\omega t +\phi)=1/2$ and on this case the acceleration would be given by;

$|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}$

And the force would be given by:

$F=ma=\frac{1}{2}mA\omega^2$

And replacing we have:

$F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N$

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3 years ago

A weightlifter lifts a 32 kg barbell a distance of 0.1 m in 0.83 s. How much power is demonstrated by the lift?

In-s [12.5K]

To find power, you have to find work first.

Given :
F = 32kg —> 313.6 N
d = 0.1m
Work = Fd
= (313.6)(0.1)
= 31.36

P=W/t
=(31.36)/(0.83)
≈ 38 W

7 0

1 year ago

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