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monitta
3 years ago
9

Is ice cream a solid -or is it actually a solid?

Physics
2 answers:
blondinia [14]3 years ago
6 0
It is a solid when is frozen and a liquid when it melts
zaharov [31]3 years ago
6 0
It has to be a liquid turned into a solid because if heat accurs then it turns back into a liquid
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“Which type of paper, construction paper or notebook paper, will make a paper airplane travel farther?”
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Answer:

Notebook paper makes paper airplane fly farther

Explanation:

It's because paper airplane made of notebook is lighter and can fly far.

4 0
1 year ago
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A circuit with two or more branches for current to flow
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If there's any point in a circuit where the current has a choice
of which branch to take, then you have a <em>parallel circuit</em>.


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3 years ago
What was anton van leeuwenhoek famous for
monitta

Answer:

He is known as the first microbiologist and also “the Father of Microbiology” because he was the first to observe bacteria underneath a microscope. He made many other significant discoveries in the field of biology and also made important changes to the microscope.

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3 years ago
Your cat "Ms." (mass 7.00 {\rm kg}) is trying to make it to the top of a frictionless ramp 2.00 {\rm m} long and inclined upward
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Final velocity at the top of the ramp is 6.58m/s

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4 0
3 years ago
When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin
dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) 1.21\times 10^{15}Hz

Explanation:

We have given Wavelength of the light  \lambda = 240 nm

According to plank's rule ,energy of light

E = h\nu = \frac{hc}{}\lambda

E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

E = KE_{max}+\Phi _{0}, here \Phi _0 is work function.

\Phi _{0}=E - KE_{max}= 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}

\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }

\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm

Part (C) In this part we have to find the cutoff frequency

\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz

5 0
3 years ago
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