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2 years ago

A sound wave of length 45 um travels with a speed of 1800 m/s in a metallic medium. Find the frequency of the sound wave.

Physics

1 answer:

Lapatulllka [165]2 years ago

5 0

Hi there!

We can use the following equation:

$\large\boxed{v = \lambda f}$

v = velocity of wave (m/s)
λ = wavelength (μm)

f = frequency (Hz)

We can rearrange the equation to solve for frequency:
$\large\boxed{f = \frac{v}{\lambda}}$

Plug in the givens. Make sure to convert μm to m:

1 μm = 0.000001 m

$f = \frac{1800}{0.000045} = 40000000 Hz = \boxed{40 \cdot 10^6 Hz}$

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A cyclist rides 6.2 km east, then 9.28 km in a direction 27.27 degrees west of north, then 7.99 km west. A. What is the magnitud

Pani-rosa [81]

Answer:

Explanation:

given,

cyclist ride 6.2 km east and then 9.28 km in the direction of 27.27° west of north and then 7.99 km west.

vertical component = 9.28 cos∅

= 9.28 cos 27.27°

= 8.24 km

horizontal axis component = 9.28 sin ∅

= 9.28 sin 27.27°

= 4.5 km

distance of the final point from the origin

= 7.99 -(6.2-4.5)

= 6.29 km

displacement

$d = \sqrt{6.29^2+8.24^2}$

d = 10.37 km

b) $tan \theta = \dfrac{6.29}{8.24}$

θ = 37.36°

4 0

3 years ago

An object thrown vertically upward from the surface of a celestial body at a velocity of 36 m/s reaches a height of sequalsminu

Anarel [89]

Answer:

v = -1.8t+36

20 seconds

360 m

40 seconds

36 m/s

The object speed will increase when it is coming down from its highest height.

Explanation:

$s=-0.9t^2+36t$

Differentiating with respect to time we get

$\frac{ds}{dt}=-1.8t+36\\\Rightarrow v=-1.8t+36$

a) Velocity of the object after t seconds is v = -1.8t+36

At the highest point v will be 0

$0=-1.8t+36\\\Rightarrow t=\frac{-36}{-1.8}\\\Rightarrow t=20\ s$

b) The object will reach the highest point after 20 seconds

$s=-0.9t^2+36t\\\Rightarrow s=-0.9\times 20^2+36\times 20\\\Rightarrow s=360\ m$

c) Highest point the object will reach is 360 m

$s=-0.9t^2+36t\\\Rightarrow 360=-0.9t^2+36t\\\Rightarrow -0.9t^2+36t-360=0\\\Rightarrow -9t^2+360t-3600=0$

$\frac{-360\pm \sqrt{0}}{2\left(-9\right)}\\\Rightarrow t=20\ s$

d) Time taken to strike the ground would be 20+20 = 40 seconds

$[tex]v=u+at\\\Rightarrow v=0+0.9\times 2\times 20\\\Rightarrow v=36\ m/s$

Acceleration will be taken as positive because the object is going down. Hence, the sign changes. 2 is multiplied because the expression is given in the form of $s=ut+\frac{1}{2}at^2$

e) The velocity with which the object strikes the ground will be 36 m/s

f) The speed will increase when the object has gone up and for 20 seconds and falls down for 20 seconds. The object speed will increase when it is coming down from its highest height.

4 0

3 years ago

A charge q of magnitude 6.4 × 10^-19 coulombs moves from point A to point B in an electric field of 6.5 × 10^4 newtons/coulomb.

lana [24]

In this problem we have the electric field intensity E:

E = 6.5 × $10^4$ newtons/coulomb