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Likurg_2 [28]
3 years ago
9

Tungsten, W-181, is a radioactive isotope with a half life of 121 days. If a medical lab purchases 24 kg of W-181, how much will

be left after 1 year?
Physics
2 answers:
luda_lava [24]3 years ago
8 0

1 year = (365 / 121) = 3.02 half-lifes.  Let's call it 3 .

The amount of radioactive isotope remaining after 3 half-lifes is

(1/2) x (1/2) x (1/2) = 1/8 

A year after the medical lab received the 24 kg of W-181, 
there will still be 24 kg of stuff in the container. 
But only 3 kg of it will still be W-181.  The other 21 kg will be
whatever substances W-181 becomes when it decays.

Sadly, even the 3 kg of good stuff won't be usable anymore ...
it'll be thoroughly mixed with the 21 kg of junk.  It would be harder
and more expensive to try and separate them than to buy a new
can of pure W-181, and USE it before 7/8 of it has deteriorated.
WITCHER [35]3 years ago
7 0

3kg if you using usatestprep

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If an automobile engine delivers a power of 50.0 hp, how much time will it take for the engine to do 6.40 x 10^4 j of work? (1 h
aleksandrvk [35]

Given data

Power (P) = 50 hp,

                = 50 × 746,  we know that 1 hp = 746 W.

                = 37300 Watts      (Watt = J/s)  

Work  = 6.40 ×10⁴ J

Power is defined as rate of doing work and the unit of power is<em> Watt.</em>

Mathematically,

         Power = (Work / time)   Watts

                     = 6.40 ×10⁴ / time

          37300 W = 6.40 ×10⁴ J /time      (Where time in seconds)

         => time = Work/Power

                      = 6.40 ×10⁴/37300

                      = <em>1.74 seconds  </em>

<em>  </em><em>Therefore , the engine need 1.74 seconds to do 6.40 6.40 ×10⁴ J of work </em>

<em> </em>


7 0
3 years ago
the speed of travel of the moon around the earth, using the formula for the speed of a moving object in a circular path
Svetach [21]

Answer: 1018.26 m/s

Explanation:

Approaching the orbit of the Moon around the Earth to a circular orbit (or circular path), we can use the equation of the speed of an object with uniform circular motion:  

V=\sqrt{G\frac{M}{r}}

Where:  

V is the speed of travel of the Moon around the Earth

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant

M=5.972(10)^{24} kg is the mass of the Earth

r=384400(10)^{3} m is the distance from the center of the Earth to the center of the Moon

Solving:

V=\sqrt{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{5.972(10)^{24} kg}{384400(10)^{3} m}}

V=1018.26 m/s This is the speed of travel of the Moon around the Earth

5 0
3 years ago
In The United States how much more sugar is the average person consume each year that in 1970?
zmey [24]

In the 1970, the average American ate only 2 pounds of sugar a year. In 1970, we ate 123 pounds of sugar per year. Today, the average American consumes almost 152 pounds of sugar in one year. This is equal to 3 pounds (or 6 cups) of sugar consumed in one week!

7 0
3 years ago
If kinetic energy of a body is decreased by 10percent then decrease in linear momentum of that body​
algol [13]

Answer:

E = 1/2 M V^2 = 1/2 P V     since P = M V

E2 / E1 = P2 V2 / (P1 V1)

P2 / P1 = E2 V1 / (E1 V2) = V2^2 V1 / (V1^2 V2) = V2 / V1

E2 / E1 = V2^2 / V1^2

V2 / V1 = (E2 / E1)^1/2

V2 / V1 = (.9)1/2 = .95

The linear momentum would have to decrease by 5%

7 0
2 years ago
Matt and Anna Killian are frequent fliers on​ Fast-n-Go Airlines. They often fly between two cities that are a distance of 1575
marin [14]

Answer:

Speed of wind = 50mi/hr, Speed of plane in still air = 400mi/hr

Explanation:

Let the speed of the wind = Vw,

Speed of the plane in still air = Vsa,

The first trip the average speed of the plane = 1575mi/4.5hours = 350mi/hr

The coming trip the wind behind = 1575mi/3.5hrs = 450

Write the motion in equation form

First trip ( the plane flew into the wind)

Vaverage = Vsa - Vw

350 = Vsa - Vw

Second trip the wind was behind

450 = Vsa +Vw

Adding the two equation

800 = 2Vas

Vas = 800/2 = 400mi/hr

Substitute for Vas into equation 1

350mi/hr = 400mi/hr - Vw

Vw = 400-350 = 50mi/hr

6 0
3 years ago
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