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Over [174]
3 years ago
10

Hii! help asap. i’ll give brainliest thanks!

Physics
1 answer:
o-na [289]3 years ago
5 0

I believe it’s the mass of the box but I don’t no if I’m right

Hope this helped
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Suppose that a block is pulled 16 meters across a floor. What amount of work is done if the force used to drag the block is 22 N
gizmo_the_mogwai [7]

wouldn't it be 22 divided by 16? i got 1.4

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In which direction does gravitational force act
Sunny_sXe [5.5K]
Gravitational force pushes down on us to keep us on the ground. So it pushes in a downward direction.
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Cuales son la energias
strojnjashka [21]

Answer:

I know the answer but cannot concert it to Spanish sorry

Explanation:

In physics, energy is the quantitative property that must be transferred to an object in order to perform work on, or to heat, the object. Energy is a conserved quantity; the law of conservation of energy states that energy can be converted in form, but not created or destroyed.

6 0
4 years ago
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What are the forces that act on the ball?​
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Answer:

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4 0
3 years ago
When light with a wavelength of 238 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic ener
erastova [34]

Answer:

The wavelength is 173 nm.

Explanation:

This kind of phenomenon is known as photoelectric effect, it occurs when photons of light inside the metal surface and if they have the right amount of energy electrons absorb it and got expelled from the metal as photo electrons. The maximum kinetic energy of that photo electrons is given by the expression:

K_{max} =E_{photon} - \Phi (1)

With E the energy of the photon and Φ the work function of the material. The work function is a value characteristic of each material and is related with how much the electron is attached to the material, the energy of the photon is the Planck's constant (h=6.63\times10^{-34}) times the frequency of light (\nu) , then (1) is:

K_{max} =h\nu - \Phi (2)

The frequency of an electromagnetic wave is related with the wavelength (\lambda) by:

\nu=\frac{c}{\lambda} (3)

with c the velocity of light (c=3.0\times10^{8})

Using (3) on (2):

K_{max} =\frac{hc}{\lambda} - \Phi

Solving for \Phi:

\Phi=\frac{hc}{\lambda}-K_max=\frac{(6.63\times10^{-34})(3.0\times10^{8})}{238\times10^{-9}}-3.13\times10^{-19}

\Phi=5.23\times10^{-19} J

That's the work function of the metal we're dealing. So now if we want to know the wavelength to obtain the double of the kinetic energy we use:

2K_{max} =\frac{hc}{\lambda} - \Phi

Solving for \lambda:

\lambda = \frac{hc}{2K_{max}+\Phi}=\frac{(6.63\times10^{-34})(3.0\times10^{8})}{2(3.13\times10^{-19})+5.23\times10^{-19}}=1.73\times10^{-7}

\lambda=173 nm

3 0
4 years ago
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