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3 years ago

The mass of a golf ball is 45.9 g . if it leaves the tee with a speed of 62.0 m/s , what is its corresponding wavelength?

Physics

1 answer:

inn [45]3 years ago

4 0

The wavelength of the golf ball is 2.328×10⁻³⁴m.

All moving particles with mass have a matter wave associated with it. These matter waves are called deBroglie waves.

The deBroglie wavelength λ of a particle is given by,

$\lambda=\frac{h}{mv}$

Here, h is the Planck's constant, m is the mass of the ball and v is its velocity.

Calculate the deBroglie wavelength of the moving golf ball by substituting 6.626×10⁻³⁴J s for h, 45.9×10⁻³kg for m and 62.0 m/s for v.

$\lambda=\frac{h}{mv}\\ =\frac{6.626*10^-^3^4Js}{(45.9*10^-^3kg)(62.0m/s)} \\ =2.328*10^-^3^4m$

The wavelength of the golf ball is 2.328×10⁻³⁴m.

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When drawing ray diagrams involving thin lenses, how many rays (at a minimum) are needed show the image distance and magnificati

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3 years ago

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nekit [7.7K]

Answer:

The image is formed at a ‘distance of 16.66 cm’ away from the lens as a diminished image of height 3.332 cm. The image formed is a real image.

Solution:

The given quantities are

Height of the object h = 5 cm

Object distance u = -25 cm

Focal length f = 10 cm

The object distance is the distance between the object position and the lens position. In order to find the position, size and nature of the image formed, we need to find the ‘image distance’ and ‘image height’.

The image distance is the distance between the position of convex lens and the position where the image is formed.

We know that the ‘focal length’ of a convex lens can be found using the below formula

1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

−

Here f is the focal length, v is the image distance which is known to us and u is the object distance.

The image height can be derived from the magnification equation, we know that

Magnification=h′h=vu\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}Magnification=

′

Thus,

h′h=vu\frac{h^{\prime}}{h}=\frac{v}{u}

′

First consider the focal length equation to find the image distance and then we can find the image height from magnification relation. So,

1f=1v−1(−25)\frac{1}{f}=\frac{1}{v}-\frac{1}{(-25)}

−

(−25)

1v=1f+1(−25)=110−125\frac{1}{v}=\frac{1}{f}+\frac{1}{(-25)}=\frac{1}{10}-\frac{1}{25}

(−25)

−

1v=25−10250=15250\frac{1}{v}=\frac{25-10}{250}=\frac{15}{250}

250

25−10

250

v=25015=503=16.66 cmv=\frac{250}{15}=\frac{50}{3}=16.66\ \mathrm{cm}v=

250

=16.66 cm

Then using the magnification relation, we can get the image height as follows

h′5=−16.6625\frac{h^{\prime}}{5}=-\frac{16.66}{25}

′

=−

16.66

So, the image height will be

h′=−5×16.6625=−3.332 cmh^{\prime}=-5 \times \frac{16.66}{25}=-3.332\ \mathrm{cm}h

′

=−5×

16.66

=−3.332 cm

Thus the image is formed at a distance of 16.66 cm away from the lens as a diminished image of height 3.332 cm. The image formed is a ‘real image’.

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3 years ago

How many Oxygen (O) atoms are in the following? H2O + CO2 4 3 2

Contact [7]

There are 3 oxygen atoms

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3 years ago

Una placa de cobre a 20°C tiene unas dimensiones de 65cm x 78 cm. Encuentra el área de la placa a 400°C; Coeficiente de dilataci

ValentinkaMS [17]

Answer:

El área de la placa es aproximadamente 5102.752 centímetros cuadrados.

Explanation:

Asumamos que el cambio dimensional como consecuencia de la temperatura es pequeña, entonces podemos estimar el área de la placa de cobre en función de la temperatura mediante la siguiente aproximación:

$A_{f} = w\cdot l \cdot [1 + 2\cdot \alpha\cdot (T_{f}-T_{o})]$ (1)

Donde:

$w$ - Ancho de la placa, en centímetros.

$l$ - Longitud de la placa, en centímetros.

$\alpha$ - Coeficiente de dilatación, en $\frac{1}{^{\circ}C}$ .

$T_{o}$ - Temperatura inicial, en grados Celsius.

$T_{f}$ - Temperatura final, en grados Celsius.

Si sabemos que $w = 65\,cm$ , $l = 78\,cm$ , $\alpha = 17\times 10^{-6}\,\frac{1}{^{\circ}C}$ , $T_{o} = 20\,^{\circ}C$ and $T_{f} = 400\,^{\circ}C$ , entonces el área de la placa a la temperatura final: