Whats the kinetic energy of an object that has a mass of 12 kilograms and moves with a velocity of 10 m/s

8. Three grams of Bismuth-218 decay to 0.375 grams in one hour. What is the half-

Evgen [1.6K]

Answer: 0.333 h

Explanation:

This problem can be solved using the Radioactive Half Life Formula:

$A=A_{o}.2^{\frac{-t}{H}}$ (1)

Where:

$A=0.375 g$ is the final amount of the material

$A_{o}=3 g$ is the initial amount of the material

$t=1 h$ is the time elapsed

$H$ is the half life of the material (the quantity we are asked to find)

Knowing this, let's substitute the values and find $h$ from (1):

$0.375 g=(3 g)2^{\frac{-1h}{H}}$ (2)

$\frac{0.375 g}{3 g}=2^{\frac{-1h}{H}}$ (3)

Applying natural logarithm in both sides:

$ln(\frac{0.375 g}{3 g})=ln(2^{\frac{-1 h}{H}})$ (4)

$-2.079=-\frac{1 h}{H}ln(2)$ (5)

Clearing $H$ :

$H=\frac{-1h}{-2.079}(0.693)$ (6)

Finally:

$h=0.333 h$ This is the half-life of the Bismuth-218 isotope

4 0

3 years ago

What is 9414 divided by 18

Levart [38]

9414/18 is equal to 523

3 0

2 years ago

A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a s

tatiyna

Answer:

a. λ = 647.2 nm

b. I₀ 9.36 x 10⁻⁵

Explanation:

Given:

β = 56.0 rad , θ = 3.09 ° , γ = 0.170 mm = 0.170 x 10⁻³ m

a.

The wavelength of the radiation can be find using

β = 2 π / γ * sin θ

λ = [ 2π * γ * sin θ ] / β

λ = [ 2π * 0.107 x 10⁻³m * sin (3.09°) ] / 56.0 rad

λ = 647.14 x 10⁻⁹ m ⇒ λ = 647.2 nm

b.

The intensity of the central maximum I₀

I = I₀ (4 / β² ) * sin ( β / 2)²

I = I₀ (4 / 56.0²) * [ sin (56.0 /2) ]²

I = I₀ 9.36 x 10⁻⁵

8 0

3 years ago

How many ounces is 200 lbs converted

ddd [48]

Answer:

3,200 ounces

Explanation:

1 pound (lb) is equal to 16 ounces (oz):

i.e 1 lb = 16 oz

Given:

200 pounds

To find:

2,000 pounds as ounces.

Steps:

200 (mass) * 16 = 3,200 ounces.

Thank you!

- EE

3 0

1 year ago

Read 2 more answers

The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the

Mama L [17]

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

P sun = I sun-earth A sun-earth

where A = 2.863 10²³×m², and I  is 1360 W/m²

P sun = 2.863 10²³ × 1360

P sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

I sun-mars = P sun / A sun-mars

where P sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

I sun-mars = 3.85×10²⁶W / 6.53 × 10²³m²

I sun-mars = 589.6 ≈ 590 W/m²

I sun-mars = 590 W/m²

6 0

3 years ago