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Bumek [7]
3 years ago
14

Nuclear Reactions: Please answer numbers 4,5,6.

Physics
1 answer:
Vikentia [17]3 years ago
8 0

Answer:

4. ¹₁H + ¹³₆C —> ¹⁴₇N + γ

5. ¹₀n + ¹⁰₅B —> ⁷₃Li + ⁴₂He

6. ⁴₂He + ¹⁴₇N —> ¹⁷₈O + ¹₁H

Explanation:

4. ¹₁H + ¹³₆C —> ¹⁴₇N + __

Let ᵇₐX be the unknown.

Thus, the equation becomes:

¹₁H + ¹³₆C —> ¹⁴₇N + ᵇₐX

Next, we shall determine b, a and X. This is illustrated below:

For b:

1 + 13 = 14 + b

14 = 14 + b

Collect like terms

b = 14 – 14

b = 0

For a:

1 + 6 = 7 + a

7 = 7 + a

Collect like terms

a = 7 – 7

a = 0

Therefore,

ᵇₐX => ⁰₀X => γ

Thus, the balanced equation is

¹₁H + ¹³₆C —> ¹⁴₇N + γ

5. ¹₀n + ¹⁰₅B —> __ + ⁴₂He

Let ˣᵧA be the unknown.

Thus, the equation becomes:

¹₀n + ¹⁰₅B —> ˣᵧA + ⁴₂He

Next, we shall determine x, y and A. This can be obtained as follow:

For x:

1 + 10 = x + 4

11 = x + 4

Collect like terms

x = 11 – 4

x = 7

For y:

0 + 5 = y + 2

5 = y + 2

Collect like terms

y = 5 – 2

y = 3

Therefore,

ˣᵧA =>⁷₃A => ⁷₃Li

Thus, the balanced equation is:

¹₀n + ¹⁰₅B —> ⁷₃Li + ⁴₂He

6. ⁴₂He + ¹⁴₇N —> __ + ¹₁H

Let ᶜₑG be the unknown.

Thus, the equation becomes:

⁴₂He + ¹⁴₇N —> ᶜₑG + ¹₁H

Next, we shall determine c, e and G. This can be obtained as follow:

For c:

4 + 14 = c + 1

18 = c + 1

Collect like terms

c = 18 – 1

c = 17

For e:

2 + 7 = e + 1

9 = e + 1

Collect like terms

e = 9 – 1

e = 8

Therefore,

ᶜₑG => ¹⁷₈G => ¹⁷₈O

Thus, the balanced equation is:

⁴₂He + ¹⁴₇N —> ¹⁷₈O + ¹₁H

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Answer:

(a) V_{avg}=2.868m/s

(b) V_{avg}=5.352m/s

(c) V_{avg}=7.836m/s

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Given data

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First we need to find distance x at these time so

x(t)=1.53t²-0.0480t³

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x(2)=1.53(2)²-0.0480(2)³=5.736m

at t=4s

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For(a) Average velocity at t=0s to t=2s

The average velocity is given as

Vavg=Δx/Δt

V_{avg}=\frac{x_{f}-x_{i} }{t_{f}-t_{i}}\\  As\\x(0)=0m\\x(2)=5.736m\\V_{avg}=\frac{5.736m-0m }{2s-0s}\\V_{avg}=2.868m/s

For(b) Average velocity at t=0s to t=4s

The average velocity is given as

Vavg=Δx/Δt

V_{avg}=\frac{x_{f}-x_{i} }{t_{f}-t_{i}}\\  As\\x(0)=0m\\x(4)=21.408m\\V_{avg}=\frac{21.408m-0m }{4s-0s}\\V_{avg}=5.352m/s

For(c) Average velocity at t=2s to t=4s

The average velocity is given as

Vavg=Δx/Δt

V_{avg}=\frac{x_{f}-x_{i} }{t_{f}-t_{i}}\\  As\\x(2)=5.736m\\x(4)=21.408m\\V_{avg}=\frac{21.408m-5.736m }{4s-2s}\\V_{avg}=7.836m/s

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Give that,

The frequency range the dog can hear is 15Hz to 50,000Hz

The wavelength of sound in air at 20°C =?

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The frequency corresponding to the lower cut-off point is the lowest frequency which his 15Hz

F=15Hz

The relationship between the wavelength, speed and frequency is given as

v=fλ

Then,

λ=v/f

λ=v/f

λ=344/15

λ=22.93m

6 0
3 years ago
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