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makvit [3.9K]
3 years ago
6

Please help!! I'm mixed up on the bottom one

Physics
1 answer:
Assoli18 [71]3 years ago
4 0
I believe the bottom one is C.
:-)
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If the atomic mass of Sodium-18 is 18.02597 u, what is the binding energy?
Klio2033 [76]

Answer:

<h3>The binding energy of sodium Na=<em>5.407791×10⁹J</em></h3>

Explanation:

<h3>Greetings !</h3>

Binding energy, amount of energy required to separate a particle from a system of particles or to disperse all the particles of the system. Binding energy is especially applicable to subatomic particles in atomic nuclei, to electrons bound to nuclei in atoms, and to atoms and ions bound together in crystals.

<h2>Formula : Eb=(Δm)c²</h2><h3>where:Eb= binding energy</h3><h3> .Δm= mass defect(kg)</h3><h3> c= speed of light 3.00×10⁸ms¯¹</h3><h2 /><h3><u>Given</u><u> </u><u>values</u></h3>
  • m= 18.02597
  • c=3.00×10⁸ms¯¹

<h3><u>required </u><u>value</u></h3>
  • Eb=?

<h3><u>Solution:</u></h3>
  • Eb=(Δm)c²
  • Eb=(18.02597)*(3.00*10⁸ms¯¹
  • Eb=5.407791*10⁹J

8 0
1 year ago
Determine the ratio of the resistivity of pure water to silver?
shutvik [7]
Silver is a very good conductor, this means its resistivity is very low (from table, we can check the precise value, which is \rho_s = 1.6 \cdot 10^{-8} \Omega m).

Pure water, instead, is a very bad conductor, this means its resistivity is very high, of order of k \Omega \cdot m (10^3 \Omega m ). Even without knowing the precise value of the pure water resistivity, we can estimate the ratio between the pure water resistivity and the silver resistivity by comparing the two orders of magnitude:
r= \frac{10^3 \Omega m}{10^{-8} \Omega m}  \sim 10^{11}

Therefore, we can say that the correct answer is
3 \cdot 10^{11} : 1
3 0
3 years ago
If the magnitude of the electric field in air exceeds roughly 3 ✕ 106 N/C, the air breaks down and a spark forms. For a two-disk
Westkost [7]

Answer:

1.843 x 10^-5 C  

Explanation:

<u><em>Givens:   </em></u>

It is given that the air starts ionizing when the electric field in the air exceeds a magnitude of 3 x 10^6 N/C, which means that the max electric field can stand without forming a spark is 3 x 10^6 N/C.  

Also it is given that the radius of the disk is 50 cm, it is required to find out the max amount of charge that the disk can hold without forming spark, which means the charge that would produce the max magnitude of the electric field that air can stand without forming spark, and since we know that the electric field in between 2 disk "Capacitor" is given by the following equation  

E = (Q/A)/∈o                                (1)

Where,

Q: total charge on the disk.

A: the area of the disk.  

<u><em>Calculations:  </em></u>

We want to find the quantity of charge on the disk that would produce an electric field of 3 x 10^6 N/C, knowing the radius of the disk we can find the cross-section of the disk, thus substituting in equation (1) we find the maximum quantity of charge the disk can hold  

Q = EA∈o

   = (3 x 10^6) x (π*0.50) x (8.85 x 10^-12)  

  = 1.843 x 10^-5 C  

note:

calculations maybe wrong but method is correct

8 0
3 years ago
Which graph shows an object that is dropped?
Olegator [25]

Answer:

I'm pretty sure it's the third one where velocity goes from positive to negative

Explanation:

the positive velocity is before the object hits the ground and the negative is after

8 0
3 years ago
A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a dra
bija089 [108]

Explanation:

Terminal velocity is given by:

v_t=\sqrt{\frac{2mg}{\rho C_dA}}

Here, m is the mass of the falling object, g is the gravitational acceleration,  C_d is the drag coefficient, \rho is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is 1.28\frac{kg}{m^3}.

v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}

7 0
3 years ago
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