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Anuta_ua [19.1K]

3 years ago

5

A tuning fork is used to produce sound waves with a frequency of 514 hertz. The waves travel through the air at 343 m/s. What is

the wavelength of the sound waves?

1 answer:

Vadim26 [7]3 years ago

5 0

Answer:

0.667 m

Explanation:

Applying,

v = λf................ Equation 1

Where v = velocity of sound, λ = wave length, f = frequency

make λ the subject of the equation

λ = v/f.............. Equation 2

From the question,

Given: v = 343 m/s, f = 514 hertz

Substitute these values into equation 2

λ = 343/514

λ = 0.667 m

Hence the wavelength of the sound wave is 0.667 m

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Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at

PSYCHO15rus [73]

Answer:

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

$\mu= \frac{m_1m_2}{m_1+m_2}$

now, since both the masses have mass m

therefore,

$\mu= \frac{m^2}{2m}$

= m/2

The final K.E. of the particles is

$KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2$

Distance between two particles is d and the gravitational potential energy between them is given by

$PE_{Gravitational}= \frac{Gmm}{d}$

By law of conservation of energy we have

$KE_{initial}+KE_{final}= PE_{gravitaional}$

Now plugging the values we get

$0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}$

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

$=\sqrt{\frac{Gm}{d} }$

This the required relation between G,m and d

5 0

3 years ago

It took 1500 Newton's of force to push a car 3 meters. How much work was done

denis23 [38]

Answer:

ow much work was done? W = F xD. IN X 2m = 2;. 2. A force of 15 newtons is ... 3. It took 50 joules to push a chair 5 meters across the floor. With what force was ... was done. How far was the rock lifted? W=FXD. D=1500 = 1.5m. Answer: :.5m ... A young man exerted a force of 9,000 newtons on a stalled car, but he was.

Explanation:

3 0

3 years ago

At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?

sergey [27]

1) Initial upward acceleration: $6.0 m/s^2$

2) Mass of burned fuel: $0.10\cdot 10^4 kg$

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude $F_g = mg$ , in the downward direction, where

$m=1.9\cdot 10^4 kg$ is the rocket's mass

$g=9.8 m/s^2$ is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

$T=3.0\cdot 10^5 N$

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

$T-mg=ma$ (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

$a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2$

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

$a'=6.9 m/s^2$

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

$T-m'g=m'a'$

where

$m'$ is the new mass of the rocket

Re-arranging the equation and solving for m', we find

$m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg$

And since the initial mass of the rocket was

$m=1.9 \cdot 10^4 kg$

This means that the mass of fuel burned is

$\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg$

3 0

3 years ago

ANSWER ASAPPPPPPPPPPPP

Free_Kalibri [48]

My teacher said it was c

6 0

3 years ago

Read 2 more answers

A very thin circular disk of radius R = 20.00 cm has charge Q = 30.00 mC uniformly distributed on its surface. The disk rotates

lutik1710 [3]

Answer:

$B= 7.5*10^{-15}T$

Explanation:

The magnetic field strenght on the z-axis at a distance d from the center is,

$B= \frac{\mu_0 Q\omega R^2}{8\pi d^3}$

Our values are:

$R=20cm\\Q=30mc\\w=5rad/s\\d=2*10^3cm=20m$

Replacing,

$B= \frac{(4\pi*10^{}-7)(30*10^{-3})(5)(0.2)^2}{8\pi(20)}$

$B= 7.5*10^{-15}T$

3 0

3 years ago