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3 years ago

An ocean wave has a frequency of 9 Hz and a speed of 18 m/s. What is the wavelength of this wave?

Physics

1 answer:

leonid [27]3 years ago

7 0

Answer:

The answer is 2 because the formula is

wavelength=speed/frequency

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Need an answer pls!!! will give brainliest

solmaris [256]

Answer:

i think your answer is C

Explanation:

8 0

3 years ago

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For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is

ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀ $e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)$

where A(t)=A₀ $e^{\frac{-bt}{2m}}$

A₀ is the amplitude at t=0 and

$w'$ is the angular frequency of damped SHM, which is given by,

$w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

Now coming to the problem,

Given: m=1.2 kg

k=9.8 N/m

b=210 g/s= 0.21 kg/s

A₀=13 cm

a) A(t)=A₀/8

⇒A₀ $e^{\frac{-bt}{2m}}$ =A₀/8

⇒ $e^{\frac{bt}{2m}}=8$

applying logarithm on both sides

⇒ $\frac{bt}{2m}=ln(8)$

⇒ $t=\frac{2m*ln(8)}{b}$

substituting the values

$t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)$

b) $w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

$w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}$

$T'=\frac{2\pi}{w'}$ , where $T'$ is time period of damped SHM

⇒ $T'=\frac{2\pi}{2.86}=2.2s$

let $n$ be number of oscillations made

then, $nT'=t$

⇒ $n=\frac{24}{2.2}=11(approx)$

8 0

3 years ago

What is a limited resource?

Artemon [7]

Something that there is only so many of or not enough of

7 0

3 years ago

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wo fixed charges, A and B are located at x axis. A is at x = 0 m, B is at x = 4 m. QA = +4.0 μC and QB = -5.0 μC. Calculate the

lys-0071 [83]

Answer:

10250 N/C leftwards

Explanation:

QA = 4 micro Coulomb

QB = - 5 micro Coulomb

AP = 6 m

BP = 2 m

A is origin, B is at 4 m and P is at 6 m .

The electric field due to charge QA at P is EA rightwards

$E_{A}=\frac{KQ_{A}}{AP^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{6^{2}}=1000 N/C (rightwards)$

The electric field due to charge QB at P is EB leftwards

$E_{B}=\frac{KQ_{B}}{BP^{2}}=\frac{9\times10^{9}\times5\times10^{-6}}{2^{2}}=11250 N/C (leftwards)$

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

5 0

3 years ago

The kicker now kicks the ball with the same speed as in the number of 4,but at 60.0°from the horizontal or 30.0° from the vertic

stealth61 [152]

Answer:

Explanation:

1) The time of flight equation for projectile motion can be used here to find total time in air.

t = 2vsin∅ / g

where v is speed, Ф is launch angle

t = 2×4×sin 60 / 9.8

t = 0.71 seconds

2) Distance where it hit the ground is called as range and has the following standard equation

D = v² sin2Ф/g

D = 4²sin 2×60 / 9.8

D = 1.41m

3) Maximum elevation is maximum time reached

h = v² sin²Ф / 2g

h = 4²sin² 60 / 2*9.8

h = 0.61 m

3 0

3 years ago