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3 years ago

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HELP ME WITH THIS ONE PLEASE!!!!

1 answer:

tensa zangetsu [6.8K]3 years ago

5 0

Answer:

?

Explanation:

cant help you if there is no question.

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A sound wave travels at 379 m/sec and has a wavelength of 8 meters. Calculate its frequency and period.

olga nikolaevna [1]

$\qquad\qquad\huge\underline{{\sf Answer}}$

Wavelength, Wavespeed and frequency depends on each other in the following way :

$\qquad \sf \dashrightarrow \: frequency = \dfrac{wave \: \: velocity}{wavelength}$

$\qquad \sf \dashrightarrow \: f = \dfrac{389}{8}$

$\qquad \sf \dashrightarrow \: f \approx 48.62 \: \: hertz$

And we know the Reciprocal relationship between frequency and period ~

So, let's find it's period •

$\sf{\qquad \sf \dashrightarrow \: t = \dfrac{8}{389}}$

$\sf{\qquad \sf \dashrightarrow \: t \approx 0.02 sec }$

6 0

2 years ago

A simple pendulum, 1.00 m in length, is released from rest when the support string is at an angle of 35.0 from the vertical. Wh

taurus [48]

Answer:

Explanation:

Length = 1.00 m

If the length is 1.0, the vertical distance pivot to bob is cos 35 = 0.819

At the lowest point, vertical distance is 1.0, so the change is the difference, 0.181 meter

The potential energy of that height is converted to kinetic energy of motion, which determines the speed.

PE = KE

mgh = ½mV²

V = √(2gh) = 1.88 m/s

7 0

3 years ago

What two elements are named after Dimitri Mendeleev?

bezimeni [28]

I'm pretty sure there is only one element named after Mendeleev: <span>Mendelevium.</span>

8 0

3 years ago

Read 2 more answers

A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a

stira [4]

Answer:

Explanation:

Given

acceleration is given by

$a=-g-c_Dv^2$

where $\ddot{y}=a$

$\dot{y}=v$

Also acceleration is given by

$a=v\frac{\mathrm{d} v}{\mathrm{d} s}$

$ds=\frac{v}{a}dv$

$\int ds=\int \frac{v}{-g-0.001v^2}dv$

$\Rightarrow Let -g-0.001v^2=t$

$-0.001\times 2vdv=dt$

$vdv=-\frac{dt}{0.002}$

$at\ v_0=50\ m/s,\ t=-g-0.001(50)^2$

$t=-g-2.5$

at $v=0,\ t=-g$

$\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}$

$\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}$

$s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}$

$s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})$

$s=113.608\ m$

when air drag is neglected maximum height reached is

$h=\frac{v_0^2}{2g}$

$h=\frac{50^2}{2\times 9.8}$

$h=127.55\ m$

3 0

3 years ago

How much force is needed to accelerate a 1,100 kg car at a rate of 1.5 m/s2?

Anna [14]

Assuming there is no force of friction...

F = ma
F = (1300kg)(1.5m/s^2)
F = 1950N
Just multiply mass by acceleration.
1300 x 1.5 = 1950N.

7 0

3 years ago