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MArishka [77]
3 years ago
5

John rode 3,150 m at an average speed of 350 m/min. If he

Physics
1 answer:
Vikentia [17]3 years ago
6 0

Answer:

mabye 8

Explanation:

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Atomic theory that states that atoms have three fundamental parts, and that electrons orbit the nucleus:
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_______ describes gases as small, energetic particles moving around and bouncing into each other?
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D. The Kinetic theory of matter
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Name two physical properties that characterize matter
STatiana [176]

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Matter have two essential types of properties, those are physical properties and chemical properties.

Explanation:

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3 years ago
An object is thrown upward with some velocity. If the object rises 77.5 m above the point of release, (a) how fast was the objec
jolli1 [7]

Answer:

v_o=39\ m/s\\t_m=4\ s

Explanation:

<u>Vertical Launch Upwards</u>

In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

\displaystyle h_m=H+\frac{v_o^2}{2g}

The object referred to in the question is thrown from a height H=0 and the maximum height is hm=77.5 m.

(a)

To find the initial speed we solve for vo:

\displaystyle v_o=\sqrt{2gh_m}

v_o=\sqrt{2\cdot 9.8\cdot 77.5}

v_o=39\ m/s

(b)

The maximum time or the time taken by the object to reach its highest  point is calculated as follows:

\displaystyle t_m=\frac{v_o}{g}

\displaystyle t_m=\frac{39}{9.8}

t_m=4\ s

7 0
3 years ago
what is the energy (in eV units) carried by one photon violet light that has a wavelength of 4.5e-7?
DaniilM [7]
The energy of a photon is given by
E=hf
where h is the Planck constant and f is the photon frequency.

We can find the photon's frequency by using the following relationship:
f= \frac{c}{\lambda}
where c is the speed of light and \lambda is the photon's wavelength. By plugging numbers into the equation, we find
f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{4.5 \cdot 10^{-7} m}=6.67 \cdot 10^{14}Hz

And so now we can find the photon energy
E=hf=(6.6 \cdot 10^{-34} Js)(6.67 \cdot 10^{14}Hz )=4.4 \cdot 10^{-19} J

We know that 1 Joule corresponds to
1 J = 1.6 \cdot 10^{-19} eV
So we can convert the photon's energy into electronvolts:
E= \frac{4.4 \cdot 10^{-19} J }{1.6 \cdot 10^{-19} J/eV}=2.75 eV
4 0
4 years ago
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