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3 years ago

How do dog whistles work?

1 answer:

7 0

<em>The sound it emits comes from what is known as the ultrasonic range, a pitch that is so high humans can't hear it. Dogs can hear these sounds, however, as can cats and other animals. Because of this, the dog whistle is a favored training tool, though it may not be for every dog parent.</em>

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In what way could a random mutation provide an organism with an advantage? With a example please

saveliy_v [14]

Answer:

They are called beneficial mutations. They lead to new versions of proteins that help organisms adapt to changes in their environment. Beneficial mutations are essential for evolution to occur. They increase an organism's changes of surviving or reproducing, so they are likely to become more common over time.

Explanation:

7 0

2 years ago

Interactions of current carrying wires

koban [17]

Physics stack exchange

7 0

3 years ago

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.320 mm wide. The diffraction pattern is observed

DanielleElmas [232]

Answer:

$W = 10.28\ mm$

Explanation:

Given,

Red light wavelength = 633 nm

width of slit = 0.320 mm

distance,d = 2.60 m

Condition of first maximum

$a sin \theta_1 = m\lambda$

$\theta_1 =sin^{-1}(\dfrac{m\lambda}{a})$

m = 1

$\theta_1 =sin^{-1}(\dfrac{633\times 10^{-9}}{0.32\times 10^{-3}})$

$\theta_1 = 0.1133^\circ$

Width of the first minima

$y_1 = L tan \theta_1$

$y_1 = 2.60\times tan( 0.11331)$

$y_1 = 5.14 \ mm$

Now, width of the central region

$W = 2 y_1$

$W = 2\times 5.14$

$W = 10.28\ mm$

8 0

3 years ago

A chipmunk with a mass of 2 kg sits on a branch that is 10 m off the ground.

vlada-n [284]

I hope I'm not too late.

GPE = mass * gravity * height

GPE = 2 kg * 9.8 m/s * 10

GPE = 196 Joules

3 0

3 years ago

Question 1 of 4 Attempt 4 The acceleration due to gravity, ???? , is constant at sea level on the Earth's surface. However, the

Evgen [1.6K]

Answer:

g(h) = g ( 1 - 2(h/R) )

<em>*At first order on h/R*</em>

Explanation:

Hi!

We can derive this expression for distances h small compared to the earth's radius R.

In order to do this, we must expand the newton's law of universal gravitation around r=R

Remember that this law is:

$F = G \frac{m_1m_2}{r^2}$

In the present case m1 will be the mass of the earth.

Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):

$F = m_2a$

Therefore, we can see that

$a(r) = G \frac{m_1}{r^2}$

With a the acceleration due to the earth's mass.

Now, the taylor series is going to be (at first order in h/R):

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R}$

a(R) is actually the constant acceleration at sea level

and

$a(R) =G \frac{m_1}{R^2} \\ \frac{da(r)}{dr}_{r=R} = -2 G\frac{m_1}{R^3}$

Therefore:

$a(R+h) \approx G\frac{m_1}{R^2} -2G\frac{m_1}{R^2} \frac{h}{R} = g(1-2\frac{h}{R})$

Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R} + \frac{h^2}{2!} \frac{d^2a(r)}{dr^2}_{r=R}$

6 0

3 years ago

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