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stepladder [879]
3 years ago
6

Do you think copper would have oxidized of it was completly submerged in vinigar

Physics
1 answer:
cluponka [151]3 years ago
5 0

yes if it was submerged long enough


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What is formed when the rock making up a tectonic plate breaks and moves Select one of the options below as your answer:
dem82 [27]

A. a fault hopes this helps you guys

6 0
3 years ago
Describe how the number of photoelectrons emitted from a metal plate in the photoelectric effect would change if the following o
andrew-mc [135]

Answer:

a) True. The number of photoelectrons is proportional to the amount (intensity) of the incident beam. From the expression above we see that threshold frequency cannot emit electrons.

b)   λ = c / f

Therefore, as the wavelength increases, the frequency decreases and therefore the energy of the photoelectrons emitted,

c)  threshold energy

        h f =Ф

Explanation:

It's photoelectric effect was fully explained by Einstein by the expression

       Knox = h f - fi

Where K is the kinetic energy of the photoelectrons, f the frequency of the incident radiation and fi the work function of the metal

a) True. The number of photoelectrons is proportional to the amount (intensity) of the incident beam. From the expression above we see that threshold frequency cannot emit electrons.

b) wavelength is related to frequency

         λ = c / f

Therefore, as the wavelength increases, the frequency decreases and therefore the energy of the photoelectrons emitted, so there is a wavelength from which electrons cannot be removed from the metal.

c) As the work increases, more frequency radiation is needed to remove the electrons, because there is a threshold energy

        h f =Ф

7 0
3 years ago
How does an atom develop a positive charge
Dmitry [639]
<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ A normal atom has the same amount of electrons and protons, making it neutral. An atom develops a positive charge when it loses an electron(s). Once it loses an electron(s), there would now be more protons that electrons.

Short answer: by losing an electron(s)

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

8 0
3 years ago
Joe has a mass of 40 kg, and Sally has a mass of 25 kg. They sit in class with a separation of 6 meters. Find the gravitational
goldfiish [28.3K]

Answer:

hmmmmmmmm

Explanation:

mmmmmmmmmmmmmmmmmmmmmmm  pay attention in class kid

4 0
3 years ago
Help with this physics task pls
cupoosta [38]

Answer:

Answers can be seen below

Explanation:

First we must explain the essential when we clear equations, and that is that if the term we need to clear is accompanied by other terms that are being added up, then those terms go to the other side of the equation to subtract if those terms are subtracting, then they go to the other side to add, if those terms are found multiplying then they go to the other side of the equation to divide and if those other terms are found dividing then they go to the other side of the equation to multiply.

(Primero debemos explicar lo esencial cuando despejamos ecuaciones, y es que si el término que necesitamos despejar va acompañado de otros términos que se están sumando, entonces esos términos van al otro lado de la ecuación para restar si esos términos están restando, luego van al otro lado para sumar, si esos términos se encuentran multiplicando luego van al otro lado de la ecuación a dividir, y si esos términos se encuentran dividiendo, pasan al otro lado de la ecuación a multiplicar.)

1 )  

t=\frac{v}{a} ; d=s*(t-t_{0} )

2)

k=\frac{2*U}{x^{2} }; T_{2}=\frac{P_{2}*V_{2}*T_{1}  }{P_{1}*V_{1}  }  \\

3)

L=\frac{F}{\pi*r*P}; d=\frac{w}{F*cos(o)}

4)

t^{2}=\frac{2*x}{g}  ; V_{2}=\frac{A_{1}*V_{1} }{A_{2} }  \\

5)

h=\frac{V}{\pi *r^{2} } ; r=\frac{t}{F*sin(o)}

6)

h=\frac{m}{(1/2)*\pi *r^{2} }  ; h_{2}=\frac{F_{2}*(1/2)*b_{1} *h_{1} }{F_{1}*(1/2)*b_{2}*h_{2}   }

7)

b=\frac{mg-ma}{v}; m=\frac{F+kx}{g*cos(o)}

8)

a=\frac{v-v_{o} }{t} ; u=\frac{m_{1}+m_{2}  }{M}

9)

v_{o}=\frac{x-\frac{1}{2}*a*t^{2}  }{t}  ; F=\frac{W+uNd}{d*cos(o)}

10)

h=\frac{E-\frac{1}{2}*m*v^{2}  }{mg} ; v_{2} ^{2} = \frac{Dk-\frac{1}{2} m*v_{1}^{2}  }{\frac{1}{2}m }

11)

N=\frac{mg*sin(o)-F}{u} ; x^{2}=\frac{W+\frac{1}{2}k*x_{1}^{2}   }{\frac{1}{2}*k }

12)

x=x_{o} +\frac{v^{2-v_{o}^{2}  } }{2a}  ;  m=\frac{P*A-F_{1}-F_{2} }{g}

13)

x_{o} = x-\frac{F}{k} ;  u=\frac{cos(o)-\frac{a}{g} }{sin(o)}

14)

t=\frac{d}{v} +t_{o} ; t_{o} = t-(\frac{v-v_{o} }{a} )

15)

F_{2}=\frac{W-F_{1} *d}{d}+F_{3}   ;  v_{2}^{2}=v_{1}^{2}+\frac{2*Dk}{m}

16)

y_{1}=y-\frac{u}{mg}  ; x^{2} = \frac{2W}{k}+x_{o} ^{2}

7 0
3 years ago
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