Hey can anybody help me with this question please

The half-life of the radioactive element unobtanium-43 is 10 seconds. If 48 grams of unobtanium-43 are initially present, how ma

natka813 [3]

Answer:

At time, 10 seconds, the mass remaining is 24 grams

At time, 20 seconds, the mass remaining is 12 grams

At time, 30 seconds, the mass remaining is 6 grams

At time, 40 seconds, the mass remaining is 3 grams

At time, 50 seconds, the mass remaining is 1.5 grams

Step-by-step explanation:

Given;

initial mass of the radioactive element = 48 grams

half life, t = 10 seconds

time of decay (s) remaining mass of the radioactive element

0 ------------------------------------------------- 48 grams

10------------------------------------------------- 24 grams

20 ------------------------------------------------- 12 grams

30 -------------------------------------------------- 6 grams

40 --------------------------------------------------- 3 grams

50 ----------------------------------------------------- 1.5 grams

60------------------------------------------------------0.75 grams

Thus;

At time, 10 seconds, the mass remaining is 24 grams

At time, 20 seconds, the mass remaining is 12 grams

At time, 30 seconds, the mass remaining is 6 grams

At time, 40 seconds, the mass remaining is 3 grams

At time, 50 seconds, the mass remaining is 1.5 grams

7 0

3 years ago

Fracción equivalente de 1/3

Scrat [10]

Answer:

2/6

Step-by-step explanation:

1/3 times 2

4 0

3 years ago

Find the volume of the composite solid below. If needed,round to the nearest hundredth. Use 3.14

ZanzabumX [31]

Answer:

429.42 cm

Step-by-step explanation:

Volume of a rectangle prism:

V = whl

V = (5)(12)(7)

V = 420

Volume of a cylinder:

V = πr²h

V = (3.14)(1²)(3)

V = (3.14)(1)(3)

V = 9.42

Combined:

420 + 9.42

429.42

6 0

3 years ago

Identify the slope and y-intercept of the function y = 4x - 9. O A. The slope is -9. The yintercept is (0,4). B. The slope is 4.

RUDIKE [14]

Answer:

B

Step-by-step explanation:

6 0

3 years ago

(b) If a series follows geometric progression, show that the ratio of the sum of term to the sum from (n+1) term to (2n) term is

Nadusha1986 [10]

Proof with explanation:

We know that the sum of first 'n' terms of a Geometric progression is given by

$S_{n}=\frac{a(1-r^{n})}{1-r}$

where

a = first term of G.P

r is the common ratio

'n' is the number of terms

Thus the sum of 'n' terms is

$S_{n}=\frac{a(1-r^{n})}{1-r}$

Now the sum of first '2n' terms is

$S_{2n}=\frac{a(1-r^{2n})}{1-r}$

Now the sum of terms from $(n+1)^{th}$ to $(2n)^{th}$ term is $S_{2n}-S_{n}$

Thus the ratio becomes

$\frac{S_{n}}{S_{2n}-S_{n}}\\\\=\frac{\frac{a(1-r^{n})}{1-r}}{\frac{a(1-r^{2n})}{1-r}-\frac{a(1-r^{n})}{1-r}}\\\\=\frac{1-r^{n}}{r^{n}-r^{2n}}\\\\=\frac{1-r^{n}}{r^{n}(1-r^{n})}\\\\=\frac{1}{r^{n}}$

5 0

3 years ago