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Kay [80]
3 years ago
9

What is the wavelength associated with an electron with a velocity of 4.8X10s m/s? (Mass of the electron is 9.1X10-31kg)?

Physics
1 answer:
amid [387]3 years ago
4 0

Answer:

1.52 nm

Explanation:

Using the De Broglie wavelength equation,

λ = h/p where λ = wavelength associated with electron, h = Planck's constant = 6.63 × 10⁻³⁴ Js and p = momentum of electron = mv where m = mass of electron = 9.1 × 10⁻³¹ kg and v = velocity of electron = 4.8 × 10⁵ m/s

So, λ = h/p

λ = h/mv

substituting the values of the variables into the equation, we have

λ = h/mv

λ = 6.63 × 10⁻³⁴ Js/(9.1 × 10⁻³¹ kg × 4.8 × 10⁵ m/s)

λ = 6.63 × 10⁻³⁴ Js/(43.68 × 10⁻²⁶ kgm/s)

λ = 0.1518 × 10⁻⁸ m

λ = 1.518 × 10⁻⁹ m

λ = 1.518 nm

λ ≅ 1.52 nm

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3 years ago
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A 1000 kg satellite and a 2000 kg satellite follow exactly the same orbit around the earth. What is the ratio F1/F2 of the gravi
Tamiku [17]

Answer:

the <em>ratio F1/F2 = 1/2</em>

the <em>ratio a1/a2 = 1</em>

Explanation:

The force that both satellites experience is:

F1 = G M_e m1 / r²       and

F2 = G M_e m2 / r²

where

  • m1 is the mass of satellite 1
  • m2 is the mass of satellite 2
  • r is the orbital radius
  • M_e is the mass of Earth

Therefore,

F1/F2 = [G M_e m1 / r²] / [G M_e m2 / r²]

F1/F2 = [G M_e m1 / r²] × [r² / G M_e m2]

F1/F2 = m1/m2

F1/F2 = 1000/2000

<em>F1/F2 = 1/2</em>

The other force that the two satellites experience is the centripetal force. Therefore,

F1c = m1 v² / r    and

F2c = m2 v² / r

where

  • m1 is the mass of satellite 1
  • m2 is the mass of satellite 2
  • v is the orbital velocity
  • r is the orbital velocity

Thus,

a1 = v² / r ⇒ v² = r a1    and

a2 = v² / r ⇒ v² = r a2

Therefore,

F1c = m1 a1 r / r = m1 a1

F2c = m2 a2 r / r = m2 a2

In order for the satellites to stay in orbit, the gravitational force must equal the centripetal force. Thus,

F1 = F1c

G M_e m1 / r² = m1 a1

a1 = G M_e / r²

also

a2 = G M_e / r²

Thus,

a1/a2 = [G M_e / r²] / [G M_e / r²]

<em>a1/a2 = 1</em>

4 0
3 years ago
2. At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a n
AleksandrR [38]

Answer:

No

Explanation:

Let the reference origin be location of ship B in the beginning. We can then create the equation of motion for ship A and ship B in term of time t (hour):

A = 12 - 12t

B = 9t

Since the 2 ship motions are perpendicular with each other, we can calculate the distance between 2 ships in term of t

d = \sqrt{A^2 + B^2} = \sqrt{(12 - 12t)^2 + (9t)^2}

For the ships to sight each other, distance must be 5 or smaller

d \leq 5

\sqrt{(12 - 12t)^2 + (9t)^2} \leq 5

(12 - 12t)^2 + (9t)^2 \leq 25

144t^2 - 288t + 144 + 81t^2 - 25 \leq 0

225t^2 - 288t + 119 \leq 0

(15t)^2 - (2*15*9.6)t + 9.6^2 + 26.84 \leq 0

(15t^2 - 9.6)^2 + 26.84 \leq 0

Since (15t^2 - 9.6)^2 \geq 0 then

(15t^2 - 9.6)^2 + 26.84 > 0

So our equation has no solution, the answer is no, the 2 ships never sight each other.

8 0
3 years ago
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6 0
2 years ago
1a) An object is thrown up with a velocity of 18m/s from a height of 520m.
xeze [42]

Answer:

Explanation:

An object is thrown up from a height of 520m

h_o = 520m

Initial velocity of thrown is 18m/s

u = 18m/s

What is the position of the object after 3seconds

t = 3s

Acceleration due to gravity

g = 9.81 m/s²

Let calculated the height the object will reached when thrown from the top.

Using equation of motion

h = ut + ½gt²

Since the body is thrown upward, it is acting against gravity then, gravity will be negative

Then,

h = ut — ½gt²

h = 18 × 3 — ½ × 9.81 × 3²

h = 54 — 44.145

h = 9.855 m

So, the body is above the top of the building at a distance of 9.855m

So, the total distance from the bottom is

Position = h + h_o

x = 9.855 + 520

x = 529.855m

x ≈ 530m,

The position of objects after 3seconds is 520m

3 0
3 years ago
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