All categories

3 years ago

If I start with an 30 gram sample of Plutonium-244, how many grams would be left of Plutonium-244 after 1 half life

Physics

1 answer:

andriy [413]3 years ago

4 0

Answer:

15 g

Explanation:

Given,

Plutonium-244 = 30 g

We have to find the gram of Plutonium-244 left after 1 half-life.

We know that the half-life of a radioactive isotope is constant. The half-life of a Radioactive isotope does not depend on the initial amount of isotope.

Now,

After 1 half-life

Plutonium-244 left = $\dfrac{30}{2}$

= 15 g

Hence, 15 g of the Plutonium-244 will be left after 1 half-life.

You might be interested in

A diverging lens has a focal length of 18.0 cm. An insect is placed 7.00 cm in front of the lens. What is the magnification

musickatia [10]

Diverging lens=concave lens

The image would be after the first focal point but before the lens and it will be smaller than the object. Look at my picture for reference. Please let me know if this helped you!

3 0

3 years ago

The human body can store excess animo acids as a protein source ?

Bezzdna [24]

Answer:

The thermic effect (TEF) is greater for protein compared to carbohydrate and fat. The body can store excess amino acids for later use. The protein-sparing effect spares the utilization of protein as an energy source and allows it to be used for more important metabolic functions.

3 0

3 years ago

If the mean velocity adjacent to the top of a wing of 1.8 m chord is 40 m/s and that adjacent to the bottom of the wing is 31 m/

Cloud [144]

Answer:

lift per meter of span = 702 N/m

Explanation:

See attached pictures.

7 0

4 years ago

Far from any other masses, two masses, m1 and m2, are interacting gravitationally. The value for the mass of m1 suddenly doubles

NNADVOKAT [17]

<span>It also doubles The gravitational force between two masses is expressed as: F = G*m1*m2/r^2 where F = Force between the two masses m1 = Mass of object 1 m2 = Mass of object 2 r = distance between centers of object 1 and object 2 G = Gravitational constant The exact values of G, m1, m2, and r don't matter since all except for m1 is held constant. And when m1 suddenly doubles, the force attracting the two object to each other also doubles.</span>

5 0

4 years ago

A car travels on a banked circular curve at constant speed. All four of the car's tires roll without slipping. The car moves at

kvv77 [185]

Answer:

b. The horizontal component of the normal force acts toward the center of the circle.

Explanation:

When a car on a banked turn move with just the right speed so that the frictional force is zero, a component of the normal force -which is always perpendicular to the surface- acts as the centripetal force on the car. This is because the normal force now has a horizontal component, and this component supplies the centripetal force.

7 0

3 years ago