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3 years ago

If there are 50 grams of U-238 on day zero of radioactive decay, how much will there be after 4.5 billion years

2 answers:

4 0

We use the formula of the half-life to calculate for the remaining U-238 after 4.5 billion years. The formula is expressed as A = A₀ (1/2)^(t/h) where A is the final amount, A₀ is the initial amount of the substance, t is the time and h is the half-life of the substance wherein for U-238 h is equal to 4.47 billion years.

A = A₀ (1/2)^(t/h)
A = 50 (1/2)^(4.5 / 4.47)
A = 24.88 g

kobusy [5.1K]3 years ago

4 0

25 if using usatsyprep

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An office window has dimensions 3.1 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.954 a

Virty [35]

Answer:

Net forces which pushes the window is 30342.78 N.

Explanation:

Given:

Dimension of the office window.

Length of the window = $3.1$ m

Width of the window = $2.1$ m

Area of the window = $(3.1\times 2.1) = 6.51\ m^2$

Difference in air pressure = Inside pressure - Outside pressure

= $(1.0-0.954)$ atm = $0.046$ atm

Conversion of the pressure in its SI unit.

⇒ $1$ atm = $101325$ Pa

⇒ $0.046$ atm = $0.046\times 101325 =4660.95$ Pa

We have to find the net force.

We know,

⇒ Pressure = Force/Area

⇒ $Pressure=\frac{Force }{Area}$

⇒ $Force =Pressure\times Area$

⇒ Plugging the values.

⇒ $Force =4660.95\times 6.51$

⇒ $Force=30342.78$ Newton (N)

So,

The net forces which pushes the window is 30342.78 N.

3 0

3 years ago

How do you solve 0.004 dm + 0.12508 dm?

Effectus [21]

0.004 of something added to 0.12508 of the same thing
adds up to 0.12908 of it.

The thing could be a glass of water, a sheet of paper,
a pound of ground beef, a gallon of gas, or a snowball.
In this problem, it just happens to be a dm.

7 0

3 years ago

You drop a rock down a well that is 5.4 m deep. How long does it take the rock to hit the bottom of the well?

Natali5045456 [20]

Equation of motion:

$y_{f}=y_{o}+v_{o}t+ \frac{1}{2} at^{2}$

Since initial velocity is zero, the second term goes away:

$y_{f}=y_{o}+0+ \frac{1}{2} at^{2}$

$y_{f}=y_{o}+\frac{1}{2} at^{2}$

$y_{f}-y_{o}= \frac{1}{2} at^{2}$

$y_{f}-y_{o}=5.4m$

$5.4m= \frac{1}{2} at^{2}$

$\frac{2(5.4m) }{a} = t^{2}$

$a = g = 9.81 \frac{m}{ s^{2}}$

$\frac{2(5.4m) }{9.81 \frac{m}{ s^{2} } } = t^{2}$

$1.1 s^{2} = t^{2}$ $\sqrt{1.1 s^{2}} = \sqrt{t^{2}}$

 $t = 1.05s$ 

4 0

3 years ago

Read 2 more answers

Which tagline from a car advertisement has the BEST connotative appeal to the parents of a new teenage driver?

Nonamiya [84]

A. Advanced new safety features that get everyone home at night.

6 0

3 years ago

How to do this question

Anna71 [15]

Answer:

(a) 10 m/s

(b) 22.4 m/s

Explanation:

(a) Draw a free body diagram of the car when it is at the top of the loop. There are two forces: weight force mg pulling down, and normal force N pushing down.

Sum of forces in the centripetal direction (towards the center):

∑F = ma

mg + N = mv²/r

At minimum speed, the normal force is 0.

mg = mv²/r

g = v²/r

v = √(gr)

v = √(10 m/s² × 10.0 m)

v = 10 m/s

(b) Energy is conserved.

Initial kinetic energy + initial potential energy = final kinetic energy

½ mv₀² + mgh = ½ mv²

v₀² + 2gh = v²

(10 m/s)² + 2 (10 m/s²) (20.0 m) = v²

v = 22.4 m/s

4 0

3 years ago