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1 year ago

the half-life of iodine-131 is 8.1 days. how much time had passed if i only have one-fourth of the original sample?

Physics

1 answer:

morpeh [17]1 year ago

8 0

Answer:

16.2 days

Explanation:

Find the number of halflives:

1/2 * 1/2 = 1/4 so two halflives have passed

2 * 8.1 days = 16.2 days

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A small mailbag is released from a helicopter that is descending steadily at 3 m/s.

mario62 [17]

Answer:

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Package is 44.1 meter below helicopter

Explanation:

a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Initial velocity = 3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = 3+9.8*3 = 32.4 m/s

Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m$

Distance traveled by package = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = -3+9.8*3 = 26.4 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m$

Distance traveled by package = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

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3 years ago

a runner covers the last straigjt stretch of a race in 4 s. during that time, he speeds up from 5m/s to 9m/s.

PtichkaEL [24]

During that final period of time,
his acceleration is
(9 m/s - 5 m/s) / (4 sec) = 1 m/s² .

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A fixed-geometry supersonic inlet starts at a Mach number of 3. After starting, the cruise Mach number is 2, and the operating s

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2 years ago

A student pushes a 2.85 kg cart causing it to accelerate at a rate of 4.9 m/s squared .What amount of force must the student hav

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In order to find the force (F), you would have to use the formula for it:
F=ma
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Therefore,
F=2.85kg(4.9m/s^2)
F=13.965kg(m/s^2)
Since N=kg(m/s^2)
F=13.965N
And because the problem requires that we use only 2 significant figures,
F=13N
Therefore, the student must exert 13N of force.

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1V electrical potential

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the half-life of iodine-131 is 8.1 days. how much time had passed if i only have one-fourth of the original sample?​

the half-life of iodine-131 is 8.1 days. how much time had passed if i only have one-fourth of the original sample?