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3 years ago

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Physics

1 answer:

Burka [1]3 years ago

4 0

Answer:

Second option

Explanation:

"Uniform" pretty much means the same thing happens.

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NaOH + FeCl3* Na Cl + Fe 10H)3 balanced

zvonat [6]

3NaOH + FeCl3 → 3NaCl + Fe(OH)3

8 0

3 years ago

If y = 0.02 sin (30x – 200t) (SI units), the frequency of the wave is

leva [86]

Answer:

31.831 Hz.

Explanation:

Given:

$\rm y = 0.02\sin(30x-200 t).$

The vertical displacement of a wave is given in generalized form as

$\rm y = A\sin(kx -\omega t).$

where,

A = amplitude of the displacement of the wave.
k = wave number of the wave = $\dfrac{2\pi }{\lambda}.$
$\lambda$ = wavelength of the wave.
x = horizontal displacement of the wave.
$\omega$ = angular frequency of the wave = $\rm 2\pi f$ .
f = frequency of the wave.
t = time at which the displacement is calculated.

On comparing the generalized equation with the given equation of the displacement of the wave, we get,

$\rm A=0.02.\\k=30.\\\omega =200.\\$

therefore,

$\rm 2\pi f=200\\\\\Rightarrow f = \dfrac{200}{2\pi}=31.831\ Hz.$

It is the required frequency of the wave.

3 0

3 years ago

The force of the added water produces a torque on the dam. In a simple model, if the torque due to the water were enough to caus

Fittoniya [83]

Answer:

The appropriate response is " $\tau=\frac{1}{6} PgLh^3$ ". A further explanation is described below.

Explanation:

The torque ( $\tau$ ) produced by the force on the dam will be:

⇒ $d \tau=XdF$

On applying integration both sides, we get

⇒ $\tau = \int_{0}^{a}x pgL(h-x)dx$

⇒ $= pgL\int_{0}^{h}(h-x)dx$

⇒ $=pgL[\frac{h^3}{2} -\frac{h^3}{3} ]$

⇒ $=\frac{1}{6} PgLh^3$

8 0

3 years ago

Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube op

Juliette [100K]

Answer:

The fundamental frequency of can is 2.7 kHz.

Explanation:

Given that,

A typical length for the auditory canal in an adult is about 3.1 cm, l = 3.1 cm

The speed of sound is, v = 336 m/s

We need to find the fundamental frequency of the canal. For a tube open at only one end, the fundamental frequency is given by :

$f=\dfrac{v}{4l}\\\\f=\dfrac{336}{4\times 3.1\times 10^{-2}}\\\\f=2709.67\ Hz\\\\f=2.7\ kHz$

So, the fundamental frequency of can is 2.7 kHz. Hence, this is the required solution.

7 0

3 years ago

A vector has an x-component

9966 [12]

Explanation:

The magnitude of a vector v can be found using Pythagorean's theorem.

||v|| = √(vₓ² + vᵧ²)

||v|| = √((-309)² + (187)²)

||v|| ≈ 361

You can find the angle of a vector using trigonometry.

tan θ = vᵧ / vₓ

tan θ = 187 / -309

θ ≈ 149° or θ ≈ 329°

vₓ is negative and vᵧ is positive, so θ must be in the second quadrant. Therefore, θ ≈ 149°.

4 0

3 years ago