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3 years ago

How does the number 10 relate to light, sound, force and motion?

Physics

1 answer:

Oksana_A [137]3 years ago

8 0

Answer:

number 10 is the value of acceleration due to gravity.

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A rigid object has a mass of 18kg and a radius of gyration of 0.3m. It is initially rotating clockwise about a fixed axis at its

ElenaW [278]

Answer:

:0 i dont know

Explanation:

4 0

2 years ago

A wooden rod of negligible mass and length 80.0cm is pivoted about a horizontal axis through its center. A white rat with mass 0

algol13

Answer:

The speed of the animals is 1.64m/s.

Explanation:

Let us work with variables and call the mass of the two rats $m_1$ and $m_2$ , and the length of the rod $L$ .

Using the law of conservation of energy, which says the potential energies of the rats must equal their kinetic energies, we know that when the rod swings to the vertical position,

$m_1\frac{L}{2}g -m_2\frac{L}{2}g = \frac{1}{2}m_1v^2+\frac{1}{2}m_2v^2$

$(m_1 -m_2)g\frac{L}{2} = \frac{1}{2}(m_1+m_2)v^2$ ,

solving for $v$ , we get:

$\boxed{v = \sqrt{\frac{(m_1 -m_2)gL}{(m_1+m_2)}} }$

Putting in the values for $m_1$ , $m_2$ , $g$ , and $L$ we get:

$v = \sqrt{\frac{(0.450kg -0.220kg)(9.8m/s^2)(0.8m)}{(0.450kg+0.220kg)}}$

$\boxed{ v= 1.64m/s}$

Therefore, as the rod swings through the vertical position , the speed of the rats is 1.64 m/s.

7 0

2 years ago

Electron kinetic energies are often measured in units of electron-volts (1 eV 1.6 x 10-19 J), which is the kinetic energy of an

PolarNik [594]

Answer:

4.1 eV

Explanation:

Kinetic energy, K = 0.8 eV = 0.8 x 1.6 x 10^-19 J = 1.28 x 10^-19 J

wavelength, λ = 253.5 nm = 253.5 x 10^-9 m

According to the Einstein energy equation

$E = W_{o}+K$

Where, E be the energy incident, Wo is the work function and K is the kinetic energy.

h = 6.634 x 10^-34 Js

c = 3 x 10^8 m/s

$E=\frac{hc}{\lambda }=\frac{6.634 \times 10^{-34} \times 3 \times 10^{8}}{253.5\times 10^{-9}}=7.85 \times 10^{-19} J$

So, the work function, Wo = E - K

Wo = 7.85 x 10^-19 - 1.28 x 10^-19

Wo = 6.57 x 10^-19 J

Wo = 4.1 eV

Thus, the work function of the metal is 4.1 eV.

5 0

2 years ago

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0

2 years ago

Since when was the light we see now emanating from the quasar? Note that the distance between the Earth and the quasar is 598 Mp

lakkis [162]

Hi hi hi hi hi hi hi hi hi hi hi

8 0

2 years ago