All categories

1 year ago

If a substance has a half-life of 45 min, how many minutes will it take for 220 g of the substance to be depleted to 1.00 g?.

Physics

1 answer:

AnnZ [28]1 year ago

4 0

The time taken for the substance to decay to 1 g of its original mass is 350 mins.

The half life of a radioactive element is the time taken for the half of the element to decay to half of its original size.

N = N₀(¹/₂)^t/h

where;

N is the remaining mass
N₀ is the original mass
h is half life
t is time

1 = 220(¹/₂)^t/h

1/220 = (¹/₂)^t/h

log(1/220) = log(¹/₂)^t/h

log(1/220) = t/h [log(¹/₂)]

-2.342 = t/h (-0.301)

7.78 = t/h

7.78 x h = t

7.78 x 45 min = t

350 mins = t

Thus, the time taken for the substance to decay to 1 g of its original mass is 350 mins.

Learn more about half life here: brainly.com/question/2320811

#SPJ1

You might be interested in

The sound from a single source can reach point O by two different paths. One path is 20.0 m long and the second path is 21.0 m l

aleksandrvk [35]

Answer:

minimum frequency = 170 Hz

Explanation:

given data

One path long = 20 m

second path long = 21 m

speed of sound = 340 m/s

solution

we get here destructive phase that is path difference of minimum $\frac{\lambda}{2}$

here λ is the wavelength of the wave

so path difference will be

21 - 20 = $\frac{\lambda}{2}$

λ = 2 m

and

velocity that is express as

velocity = frequency × wavelength .............1

frequency = $\frac{340}{2}$

minimum frequency = 170 Hz

7 0

3 years ago

2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate

adoni [48]

Answer:

 » Image distance :

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

v is image distance
u is object distance, u is 10 cm
f is focal length, f is 5 cm

${ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}$

 » Magnification :

• Let's derive this formula from the lens formula:

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

» Multiply throughout by fv

${ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}$