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3 years ago

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A sound wave with a frequency of 300 hertz is traveling through a medium at a speed of 320 meters/second. What is its wavelength

?

2 answers:

melamori03 [73]3 years ago

6 0

1. The wavelength is the ratio of the wave's speed to its frequency in hertz or 1/s. This is shown below,
λ = s / f = (320 m/s) / (300 1/s) = 1.07 m
The wavelength is approximately 1.07 m.

hodyreva [135]3 years ago

6 0

A wave equation is v=flambda. v is speed, f is frequency, lambda is wavelength. So, lambda = v/f. here that's 320/300 m/s. 3.2/3 1.07 metres is it ????

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The equation Bold r (t )equals(8 t plus 9 )Bold i plus (2 t squared minus 8 )Bold j plus (6 t )Bold k is the position of a parti

KATRIN_1 [288]

Explanation:

It is given that, the position of a particle as as function of time t is given by :

$r(t)=(8t+9)i+(2t^2-8)j+6tk$

Let v is the velocity of the particle. Velocity of an object is given by :

$v=\dfrac{dr(t)}{dt}$

$v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}$

$v=(8i+4tj+6k)\ m/s$

So, the above equation is the velocity vector.

Let a is the acceleration of the particle. Acceleration of an object is given by :

$a=\dfrac{dv(t)}{dt}$

$a=\dfrac{d[8i+4tj+6k]}{dt}$

$a=(4j)\ m/s^2$

At t = 0, $v=(8i+0+6k)\ m/s$

$v(t)=\sqrt{8^2+6^2} =10\ m/s$

Hence, this is the required solution.

7 0

3 years ago

If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

$s_x=v_x_0 t$

$t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s$

Now for the vertical distance

vy_o=0

than the equation of motion becomes

$s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2$

Now using this acceleration the value of electric field is calculated as

$E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\$

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

$E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C$

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0

3 years ago

How do you convert Kg's to Newtons?

MA_775_DIABLO [31]

You don't convert kilograms to newtons. By the time you've heard of these units, you know that 'kilogram' is a unit of mass, 'newton' is a unit of force or weight, and that mass and weight are different things.

Mass and force are <u>related</u> by Newton's second law:

Force = Mass x acceleration .

From this simple formula, you can see that in order to relate a mass to a force, you need to know an acceleration. And if the acceleration changes, then the relationship between the force and the mass also changes. So there's no direct conversion.

ON EARTH ONLY, one kilogram of mass <em>weighs</em> 9.8 newtons. The acceleration that connects them is the acceleration of gravity on Earth. In other places, with different gravitational accelerations, 1 kilogram weighs more or less newtons.

But they don't convert directly. That would be like asking "How do you convert miles to miles-per-hour ?"

5 0

3 years ago

In the nuclear fission process mass is converted into energy. Determine the total mass that must be converted to energy in one y

brilliants [131]

Answer:

1) Mass that needs to be converted at 100% efficiency is 0.3504 kg

2) Mass that needs to be converted at 30% efficiency is 1.168 kg

Explanation:

By the principle of mass energy equivalence we have

$E=mc^{2}$

where,

'E' is the energy produced

'm' is the mass consumed

'c' is the velocity of light in free space

Now the energy produced by the reactor in 1 year equals

$Energy=Power\times time\\\\\therefore Energy=1\times 10^{9}\times 365\times 24\times 3600\\\\Energy=31.536\times 10^{15}Joules$

Thus the mass that is covertred at 100% efficiency is

$mass=\frac{Energy}{c^{2}}\\\\mass=\frac{31.536\times 10^{15}}{(3\times 10^{8})^{2}}\\\\mass=\frac{31.536\times 10^{15}}{9\times 10^{16}}\\\\\therefore mass=0.3504kg$

Part 2)

At 30% efficiency the mass converted equals

$mass|_{30}=\frac{mass|_{100}}{0.3}\\\\mass|_{30}=\frac{0.3504}{0.3}\\\\mass|_{30}=1.168kg$

8 0

3 years ago

A 13 cm long tendon was found to stretch 3.7 mm by a force of 12.1 N . The tendon was approximately round with an average diamet

liq [111]

Answer:

young's modulus $=\frac{stress}{strain}=\frac{0.1990\times 10^{6}}{28.46\times 10^{-3}}=6.99\times 10^6N/m^2$

Explanation:

We have given length of tendon L= 13 cm =0.13 m

Change in length $\Delta L=3.7mm=3.7\times 10^{-3}m$

Force = 12.1 N

Average diameter d =8.8 mm

So $r=\frac{d}{2}=\frac{8.8}{2}=4.4mm=4.4\times 10^{-3}m$

Area $A=\pi\times (4.4\times 10^{-3})^2=60.79\times 10^{-6}m^2$

Now stress $=\frac{force}{area}=\frac{12.1}{60.79\times 10^{-6}}=0.1990\times 10^6N/m^2$

Strain $=\frac{change\ in\ lenght}{length}=\frac{3.7\times 10^{-3}}{0.13}=28.46\times 10^{-3}$

Now young's modulus $=\frac{stress}{strain}=\frac{0.1990\times 10^{6}}{28.46\times 10^{-3}}=6.99\times 10^6N/m^2$

8 0

3 years ago