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uysha [10]
2 years ago
11

A mercury atom in the ground state absorbs 20.00 electronvolts of energy and is ionized by losing an electron. How much kinetic

energy does this electron have after the ionization?
Physics
1 answer:
sukhopar [10]2 years ago
8 0
We are given a mercury atom in the ground state which absorbs 20 eV of energy. It is then ionized by losing an electron. We need to calculate the kinetic energy that the electron has after ionization.

The initial energy is 20 eV = 20 J/C
The electron charge is = 1.60217662 × 10-19<span> coulombs

To determine the kinetic energy, we can use this equation:

KE = 20 Joules / Coulombs * </span>1.60217662 × 10-19<span> coulombs 
KE = 1.25x10^20 Joules 

Therefore, the amount of kinetic energy that the electron has after ionization is </span>1.25x10^20 Joules or 1.25x10^17 kJ. <span />
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A 2 ft x 2 ft x 2 ft box weighs 100 pounds, and the weight is evenly distributed. What is the magnitude of the minimum horizonta
Tems11 [23]

Answer:

Explanation:

Let the force required be F . It is applied at the top of the box . The box is likely to turn about a corner . Torque of this force about this corner

= F x 2

This torque will try to turn the box . On the other hand the weight which is acting at CM will create a torque about the same corner . This torque will try to prevent the box to turn around the corner.

This torque of weight

= 100 x 1

= 100 pound ft.

For equilibrium

Torque of F = torque of weight.

F x 2  = 100

F = 50 pounds .

7 0
2 years ago
Read 2 more answers
dopasuj wartości pracy z ramki do przedstawionych sytuacji ,a nastepnie wyraż te prace w dżulach uwaga jedna wartośc pracy nie b
Eduardwww [97]

Answer:

A (samolot) - 200 MJ = 200000000 J

B (dźwig) - 800 kJ = 800000 J

C (podnośnik)-1.6 kJ = 1600 J

Explanation:

Całą część pytania można znaleźć na poniższym schemacie.

Z diagramu załączonego poniżej; mamy

A - samolot lotniczy

B - dźwig

C - podnoszenie ciężarów

Wszyscy to wiemy ;

1kJ = 1000 J

1MJ = 1000000 J

Mamy cztery opcje; i.e 200 MJ, 800 kJ, 1.6 kJ  and 250 mJ

Z czterech opcji można wykluczyć 250 mJ, ponieważ jest to 0,25 J, co przedstawia bardzo niską energię w porównaniu z trzema warunkami pokazanymi na schemacie.

Więc:

A (samolot) - 200 MJ = 200000000 J

B (dźwig) - 800 kJ = 800000 J

C (podnośnik)-1.6 kJ = 1600 J

Największą pracę wykona samolot. Jest tak, ponieważ ma bardzo dużą masę i bardzo dużą prędkość. W związku z tym istnieje potrzeba wytworzenia ogromnej ilości ciepła i energii.

Z drugiej strony żuraw może podnieść ładunek o wiele większy i przewyższa ciężar ciężaru, więc praca wykonywana przez dźwig musi być zdecydowanie większa niż praca ciężarka.

3 0
2 years ago
A bottle lying on the windowsill falls off and takes 4.95 seconds to reach the ground. The distance from the windowsill to the g
Liula [17]
The distance an object falls from rest through gravity is 
                        D  =  (1/2) (g) (t²) 
           Distance  =  (1/2 acceleration of gravity) x (square of the falling time)

We want to see how the time will be affected 
if  ' D ' doesn't change but ' g ' does. 
So I'm going to start by rearranging the equation
to solve for ' t '.                                                      D  =  (1/2) (g) (t²)

Multiply each side by  2 :         2 D  =            g    t²  

Divide each side by ' g ' :      2 D/g =                  t² 

Square root each side:        t = √ (2D/g)

Looking at the equation now, we can see what happens to ' t ' when only ' g ' changes:

  -- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'

                                             and smaller 'g' ==> longer 't' .-- 

They don't change by the same factor, because  1/g  is inside the square root.  So 't' changes the same amount as  √1/g  does.

Gravity on the surface of the moon is roughly  1/6  the value of gravity on the surface of the Earth.

So we expect ' t ' to increase by  √6  =  2.45 times.

It would take the same bottle  (2.45 x 4.95) = 12.12 seconds to roll off the same window sill and fall 120 meters down to the surface of the Moon.
5 0
2 years ago
Calculate the energy released in joules when one mole of polonium-214 decays according to the following equation21484 Po --&gt;
GuDViN [60]

Answer:

ΔE = 8.77 × 10¹¹ J

Explanation:

given,

²¹⁴₈₄Po -----> ²¹⁰₈₂Pb + 42 He

Atomic masses: Pb-210 = 209.98284 amu

Po-214 = 213.99519 amu

He-4 = 4.00260 amu

1 kg = 6.022 × 10²⁶ amu;

NA = 6.022 × 10²³ mol⁻¹

c = 2.99792458 × 10⁸ m/s

energy of molecule using equation

ΔE = Δm c²

Δm is mass difference and c is speed of light

Δm = 209.98284 + 4.00260 - 213.99519

Δm = - 0.00975 amu

1 amu = 1.66 x 10⁻²⁷ kg

- 0.00975 amu = - 0.00975 x 1.66 x 10⁻²⁷ Kg

                         = -0.016185 x 10⁻²⁷ Kg

total mass = 6.022 × 10²³ x -0.016185 x 10⁻²⁷

                 = -0.097467 x 10⁻⁴ Kg

ΔE = -(0.097467 x 10⁻⁴) (3 x 10^8)²

ΔE = - 8.77 × 10¹¹

ΔE = 8.77 × 10¹¹ J

8 0
2 years ago
A cubical shell with edges of length a is positioned so that two adjacent sides of one face are coincident with the +x and +y ax
Bingel [31]

Answer:

Q = ba⁴ * ε₀

Explanation:

From Gauss's Law, we know that

flux Φ = Q / ε₀

where ε₀ = 8.85e-12 C²/N·m²

and also,

Φ = EAcosθ

The field is directed along the x-axis, so that all of the flux passes through the side of the cube at x = a. This means that θ = 0º, and thus

Φ = EAcos0

Φ = EA

E = bx² meanwhile, we are interested in the point where x = a, so we substitute and then

E = ba²

Since A = a² for the cube face, we have

Q / ε₀ = E * A

Q / ε₀ = ba² * a²

so that

Q = ba⁴ * ε₀

5 0
2 years ago
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