If the absolute pressure of gas is 550.280 kPa, its gauge pressure is

A musical note has a frequency of 512 Hz. If the wavelength of the note is 0.685 m, what is the speed of the sound of that note?

riadik2000 [5.3K]

Answer:

350.72 m/s

Explanation:

Formula for velocity of wave is;

v = fλ

Where;

v is speed

f is frequency

λ is wavelength

We are given;

f = 512 Hz

λ = 0.685 m

Thus;

v = 512 × 0.685

v = 350.72 m/s

5 0

2 years ago

a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f

harkovskaia [24]

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for $\Delta$ DAC:

$tan\theta = \frac{d}{x}$

⇒ $\theta = tan^{-1} \frac{d}{x}$

Now for the $\Delta$ BAC:

$tan\theta = \frac{d + h}{x}$

⇒ $\theta = tan^{-1} \frac{d + h}{x}$

Now, differentiating w.r.t x:

$\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}]$

For maximum angle, $\frac{d\theta }{dx}$ = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]

0 = $\frac{-(d + h)}{(d + h)^{2} + x^{2}} -\frac{-d}{x^{2} + d^{2}}$

$\frac{-(d + h)}{(d + h)^{2} + x^{2}} = \frac{{d}{x^{2} + d^{2}}$

After solving the above eqn, we get

x = $\sqrt{\frac{d}{d + h}}$

The observer should stand at a distance equal to x = $\sqrt{\frac{d}{d + h}}$

4 0

2 years ago

If an object is thrown upward with an initial velocity of 128 ft/second, then its height after t seconds is given by the follow

IrinaK [193]

Answer:

The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

Explanation:

Given that,

Initial velocity u= 128 ft/sec

Equation of height

$h = 128t-32t^2$ ....(I)

(a). We need to calculate the maximum height

Firstly we need to calculate the time

$\dfrac{dh}{dt}=0$

From equation (I)

$\dfrac{dh}{dt}=128-64t$

$128-64t=0$

$t=\dfrac{128}{64}$

$t=2\ sec$

Now, for maximum height

Put the value of t in equation (I)

$h =128\times2-32\times4$

$h=128\ ft$

(b). The number of seconds it takes the object to hit the ground.

We know that, when the object reaches ground the height becomes zero

$128t-32t^2=0$

$t(128-32t)=0$

$128=32t$

$t=4\ sec$

Hence, The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

3 0

3 years ago

Hypothetically speaking, if an object were located at the center of the Earth, the gravitational force on that object due to the

Likurg_2 [28]

Answer:

D. ) The force would be zero newtons

Explanation:

Because

If you are at the center of the earth, gravity is zero because all the mass around you is pulling "up" (every direction there is up!)

So F=mg so if g is zero F is also zero

5 0

3 years ago

A helicopter blade spins at exactly 110 revolutions per minute. Its tip is 4.50 m from the center of rotation. (a) Calculate th

NeX [460]

Answer:

(a). The average speed is 51.83 m/s.

(b). The average velocity over one revolution is zero.

Explanation:

Given that,

Angular velocity = 110 rev/m

Radius = 4.50 m

(a). We need to calculate the average speed

Using formula of average speed

$v=r\omega$

$v = 4.50\times110\times\dfrac{2\pi}{60}$

$v=51.83\ m/s$

(b). The average velocity over one revolution is zero because the net displacement is zero in one revolution.

Hence, (a). The average speed is 51.83 m/s.

(b). The average velocity over one revolution is zero.

8 0

2 years ago