Ask question

Ask question

All categories

3 years ago

11

After two hours, 1/16 of the initial amount of a certain radioactive isotope remains undecayed. The half-life of the isotope is:

1 answer:

AlekseyPX3 years ago

7 0

The half life is 30 minutes.
30 mins- 1/2 left
60 mins- 1/4 left
90 mins- 1/8 left
120 mins- 1/16 left
120 mins= 2 hours

You might be interested in

What are positive effects of plastics?

tamaranim1 [39]

<span>It is light, easily shaped, strong, and inexpensive. Its ability to guard against contamination makes it useful in sterile medical environments such as hospitals.</span>

3 0

3 years ago

Read 2 more answers

What is the current I(3τ), that is, the current after three time constants have passed? The current in the circuit will approach

Olin [163]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$I(\tau)=0.051 A$

b

$I(3 \tau)=0.076 A$

c

$I_c= 0.08 A$

Explanation:

From the question we are told that

$I(t) = \frac{e}{R}(1-e^{\frac{t}{\tau} }) ; \ Where \ \tau = L/R$

From the question we are told to find $I(\tau)$ when t=0 equals the time constant ( $\tau$ )

That is to obtain $I(\tau)$ .This is mathematically represented as

$I(\tau = t) = \frac{\epsilon}{R} (1- e^{-\frac{\tau}{\tau} })$

Substituting 12 V for $\epsilon$ and 150Ω for R

$I(\tau) = \frac{12}{150} (1- e^{-1})$

$=0.051 A$

From the question we are told to find $I(3 \tau)$ when t=0 equals the 3 times the time constant ( $\tau$ )

That is to obtain $I(3\tau)$ .This is mathematically represented as

$I(\tau = t) = \frac{\epsilon}{R} (1- e^{-\frac{3\tau}{\tau} })$

$I(\tau) = \frac{12}{150} (1- e^{-3})$

$=0.076 A$

As tends to infinity $\frac{\infty}{\tau} = \infty$

So $I_c$ would be mathematically evaluated as

$I_c=I(\infty) = \frac{12}{150} (1- e^{- \infty})$

$= \frac{12}{150}$

$= 0.08 A$

5 0

3 years ago

A roller coaster is released from the top of a track that is 57 m high. Find the roller coaster speed when it reaches ground lev

Art [367]

Hi there!

Assuming the track is frictionless:

$E_i = E_f\\\\PE = KE\\\\mgh = \frac{1}{2}mv^2$

Cancel out the masses and rearrange to solve for velocity:

$gh = \frac{1}{2}v^{2}\\\\v = \sqrt{2gh}\\\\$

Plug in the given height and let g = 9.8 m/s²:

$v = \sqrt{2(9.8)(57)} = \boxed{33.42 m/s}$

7 0

3 years ago

A 75.0 kg driver slides into a seat in a truck that has only 1 spring. The spring compresses 1.60 cm = 0.0160 m. What is the spr

ValentinkaMS [17]

Answer

0.510

Explanation:

5 0

3 years ago

Expresa en terminos de la unidad fundamental

Svet_ta [14]

Answer:

imma be honest I dint really know

5 0

3 years ago