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3 years ago

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A tuning fork has a frequency of 280 Hz and the wavelength of the sound produced is 1.5 meters. Calculate the wave's frequency a

nd period. Show Your Work!

1 answer:

s2008m [1.1K]3 years ago

3 0

Answer:

The answer would be 420 m/s

Explanation:

Look in attachment ⬇

I Hope this Helps!!!

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Suppose you have two meter sticks, one made of steel and one made of invar (an alloy of iron and nickel), which are the same len

Mekhanik [1.2K]

Answer:

The difference in length for steel is 2.46 x 10⁻⁴ m

The difference in length for invar is 1.845 x 10⁻⁵ m

Explanation:

Given;

original length of steel, L₁ = 1.00 m

original length of invar, L₁ = 1.00 m

coefficients of volume expansion for steel, $\gamma_{st.}$ = 3.6 × 10⁻⁵ /°C

coefficients of volume expansion for invar, $\gamma_{in.}$ = 2.7 × 10⁻⁶ /°C

temperature rise in both meter stick, θ = 20.5°C

Difference in length, can be calculated as:

L₂ = L₁ (1 + αθ)

L₂ = L₁ + L₁αθ

L₂ - L₁ = L₁αθ

ΔL = L₁αθ

Where;

ΔL is difference in length

α is linear expansivity = $\frac{\gamma}{3}$

Difference in length, for steel at 20.5°C:

ΔL = L₁αθ

Given;

L₁ = 1.00 m

θ = 20.5°C

$\alpha = \frac{\gamma}{3} = \frac{3.6*10^{-5}}{3} = 1.2*10^{-5} /^oC$

ΔL = 1 x 1.2 x 10⁻⁵ x 20.5 = 2.46 x 10⁻⁴ m

Difference in length, for invar at 20.5°C:

ΔL = L₁αθ

Given;

L₁ = 1.00 m

θ = 20.5°C

$\alpha = \frac{\gamma}{3} = \frac{2.7*10^{-6}}{3} = 0.9*10^{-6}/^oC$

ΔL = 1 x 0.9 x 10⁻⁶ x 20.5 = 1.845 x 10⁻⁵ m

8 0

3 years ago

Suppose that you'd like to find out if a distant star is moving relative to the earth. The star is much too far away to detect a

rjkz [21]

Answer:

The speed will be "18km/s". A further explanation is given below.

Explanation:

According to the question, the values are:

Wavelength,

$\lambda = 656.46 \ nm$

$\Delta \lambda = 0.04$

$c=3\times 10^8$

As we know,

⇒ $\frac{\Delta \lambda}{\lambda} =\frac{v}{c}$

On substituting the values, we get

⇒ $\frac{656.46}{0.04} =\frac{v}{3\times 10^8}$

⇒ $v=\frac{656.46}{0.04} (3\times 10^8)$

⇒ $=16411.5\times 3\times 10^8$

⇒ $=18280 \ m/s$

or,

⇒ $=18 \ km/s$

8 0

3 years ago

A football is kicked from the ground with a velocity of 38m/s at an angle of 40 degrees and eventually lands at the same height.

Anastasy [175]

Initially, the velocity vector is $\langle 38cos(40^{\circ}),38sin(40^{\circ}) \rangle=\langle 29.110, 24.426 \rangle$ . At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by $4.9(0.2)^2$ , so the velocity is $\langle 29.110, 24.426-0.196 \rangle = \langle 29.110, 24.23 \rangle$ .

Converting back to direction and magnitude, we get $\langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle$

4 0

3 years ago

In a double slit experiment, if the separation between the two slits is 0.050 mm and the distance from the slits to a screen is

meriva

The spacing between the first-order and second-order bright fringes is 3 cm. Hence, this is the required solution.

<h3>What is double slit experiment?</h3>

The double-slit experiment serves as a proof in current physics that both light and matter may exhibit properties of classically defined waves and particles. It also illustrates the inherently probabilistic nature of quantum mechanical events. Thomas Young carried out the first experiment of this kind employing light in 1802, illustrating how light behaves like a wave. It was formerly believed that light was made up of either waves or particles. About a century later, with the advent of modern physics, it was discovered that light may in fact exhibit behaviour like that of both waves and particles. The identical behaviour of electrons was first shown by Davisson and Germer in 1927, and it was later extended to atoms and molecules.

The separation between the slits, d = 0.05mm = 5×10⁻⁵ m

The distance from the slits to a screen, D = 2.5 m

Let x is the spacing between the first-order and second-order bright fringes when coherent light of wavelength 600 nm illuminates the slits,

λ = 600nm = 6× 10⁻⁷ m

We know that the bright fringe is given by :

y = nλD/d

So, the spacing between the first-order and second-order bright fringes is :

x = 2λD/d - λD/d

x = λD/d

x = 6 × 10⁻⁷ × 2.5/5 × 10⁻⁵

x = 0.03 m

or

x = 3 cm

So, the spacing between the first-order and second-order bright fringes is 3 cm. Hence, this is the required solution.

to learn more about double slit experiment go to -

brainly.com/question/28108126

#SPJ4

8 0

1 year ago

An object has 1200 J of gravitational potential energy when it is dropped. Right when the object hits the ground how much kineti

Fittoniya [83]

Answer:

The kinetic energy is 1200 J

Explanation:

The Principle of Conservation of energy states that "energy is neither created nor destroyed, it is transformed".

This means that energy can be transformed from one form to another, but the total amount of energy always remains constant, that is, the total energy is the same before and after each transformation.

The mechanical energy of a body or a physical system is the sum of its kinetic energy and the potential energy. According to the Principle of Conservation of Energy for mechanical energy, the total mechanical energy that a body possesses is constant at every instant of time.

Since mechanical energy is equal to the sum of kinetic energy and gravitational potential energy that a body possesses, the only way to stay constant is that:

when the kinetic energy increases the gravitational potential energy decreases,
when gravitational potential energy increases, kinetic energy decreases.

Due to the Principle of Conservation of Energy you can say that the gravitational potential energy is converted to kinetic energy. So Gravitational potential energy at the top = kinetic energy at the bottom

<u><em>The kinetic energy is 1200 J</em></u>

5 0

3 years ago