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3 years ago

How many picoseconds are there in 1 Ms- show work

Physics

1 answer:

Naily [24]3 years ago

7 0

There are 10⁹ picoseconds in 1 Ms

1 picosecond= 10¹² s

1 Ms = 10⁻³ s

so the number of picoseconds in one Ms=(10⁻³ s/1 Ms) * (10¹² Ps/ 1 s)=10⁹

Thus there are 10⁹ picoseconds in 1 Ms

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Answer:

Explanation:burn the gas

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3 years ago

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3 years ago

What is the electric force acting between two charges of -0.0050 C and 0.0050 C that are 0.025 m apart?

nika2105 [10]

Answer:

(A) $-3.6\times 10^8\ N$

Explanation:

Given:

Charge of one particle (q₁) = -0.0050 C

Charge of another particle (q₂) = 0.0050 C

Separation between them (d) = 0.025 m

We know that, from Coulomb's law, electric force acting between two charged particles is given as:

$F_e=\dfrac{kq_1q_2}{d^2}\\\\Where,k\to Coulomb's\ constant = 9\times 10^9\ N\cdot m^2/C^2$

Plug in the given values and solve for electric force, $F_e$ . This gives,

$F_e=\frac{(9\times 10^9\ N\cdot m^2/C^2) (-0.0050\ C)(0.0050\ C)}{(0.025\ m)^2}\\\\F_e=\frac{-2.25\times 10^{-4}\times 10^9}{6.25\times 10^{-4}}\ N\\\\F_e=-3.6\times 10^8\ N$

Therefore, option (A) is correct. Negative sign implies that the nature of electric force is attraction.

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3 years ago

Please choose the answer that describes the scientific notation for 5,098,000.

galben [10]

D. 5.098 x 106 is the correct answer when reduced to the proper notation.

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4 years ago

A sample contains radioactive atoms of two types, A and B. Initially there are five times as many A atoms as there are B atoms.

victus00 [196]

Answer:

Explanation:

Initially no of atoms of A = N₀(A)

Initially no of atoms of B = N₀(B)

5 X N₀(A) = N₀(B)

N = N₀ $e^{-\lambda t}$

N is no of atoms after time t , λ is decay constant and t is time .

For A

N(A) = N(A)₀ $e^{-\lambda_1 t}$

For B

N(B) = N(B)₀ $e^{-\lambda_2 t}$

N(A) = N(B) , for t = 2 h

N(A)₀ $e^{-\lambda_1 t}$ = N(B)₀ $e^{-\lambda_2 t}$

N(A)₀ $e^{-\lambda_1 t}$ = 5 x N₀(A) $e^{-\lambda_2 t}$

$e^{-\lambda_1 t}$ = 5 $e^{-\lambda_2 t}$

$e^{\lambda_2 t}$ = 5 $e^{\lambda_1 t}$

half life = .693 / λ

For A

.77 = .693 / λ₁

λ₁ = .9 h⁻¹

$e^{\lambda_2 t}$ = 5 $e^{\lambda_1 t}$

Putting t = 2 h , λ₁ = .9 h⁻¹

$e^{\lambda_2\times 2}$ = 5 $e^{.9\times 2}$

$e^{\lambda_2\times 2}$ = 30.25

2 x λ₂ = 3.41

λ₂ = 1.7047

Half life of B = .693 / 1.7047

= .4065 hours .

= .41 hours .

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3 years ago