answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
babymother [125]
2 years ago
10

A radioactive substance decays exponentially. A scientist begins with 200 milligrams of a radioactive substance. After 17 hours,

100 mg of the substance remains.How many milligrams will remain after 24 hours?
Physics
1 answer:
lianna [129]2 years ago
3 0

75.17 mg of the radioactive substance will remain after 24 hours.

Answer:

Explanation:

Any radioactive substance will obey the exponential decay behavior. So according to this behavior, any radioactive substance will be decaying in terms of exponential form of disintegration constant and Time.

Disintegration constant is the rate of decay of radioactive elements. It can be measured using the half life time of the radioactive element .While half life time is the time taken by any radioactive element to decay half of its concentration. Like in this case, at first the scientist took 200 mg then after 17 hours, it got reduced to 100 g. So the half life time of this element is 17 hours.

Then Disintegration constant = 0.6932/Half Life time

Disintegration constant = 0.6932/17=0.041

Then as per the law of disintegration constant:

N = N_{0}e^{-xt}

Here N is the amount of radioactive element remaining at time t and N_{0} is the initial amount of sample, x is the disintegration constant.

So here, N_{0} = 200 mg, x = 0.041 and t = 24 hrs.

N = 200 ×e^{-24*0.041} =75.17 mg.

So 75.17 mg of the radioactive substance will remain after 24 hours.

You might be interested in
A toy car is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.0 m/s656-03-02-00-00_files/i02900
Whitepunk [10]
It would be 17 m/s

If we use

V2 = V1 + a*t
Sub in 5 for v1
2m/s*2 for a
And
6 for t
That should give you the answer.
5 0
2 years ago
Read 2 more answers
Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for
Burka [1]

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

3 0
2 years ago
Which pair of sentences is describing the same velocity? A car is parked. A car is moving in circles. A bus drives 40 miles per
olya-2409 [2.1K]
The bus and the truck have the same velocity.

Also, Valerie and Owen have the same velocity.
8 0
2 years ago
Read 2 more answers
g A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobb
Law Incorporation [45]

Answer:

Explanation:

total weight acting downwards

= 3g + 10g

13 g

volume of lead = 10 / 11.3 = .885 cm³

Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber  = v x 1 x g

Buoyant force on lead =  .885 x 1 x g

total buoyant force = vg + .885 g

For floating

vg + .885 g  = 13 g

v = 12.115 cm³

total volume of bobber

= 4/3 x 3.14 x 2³

= 33.5 cm³

fraction of volume submerged

= 12.115  / 33.5

= .36  

= 36 %

4 0
2 years ago
How much heat Q1 is transferred by 25.0 g of water onto the skin? To compare this to the result in the previous part, continue t
hodyreva [135]

Answer:

The heat transferred  from water to skin  is 6913.5 J.

Explanation:

Given that,

Weight of water = 25.0 g

Suppose that water and steam, initially at 100°C, are cooled down to skin temperature, 34°C, when they come in contact with your skin. Assume that the steam condenses extremely fast. We will further assume a constant specific heat capacity c=4190 J/(kg°K) for both liquid water and steam.

We need to calculate the heat transferred  from water to skin

Using formula for stream

Q=mc\Delta T

Put the value into the formula

Q=25\times10^{-3}\times4190\times(373-307)

Q=6913.5\ J

Hence, The heat transferred  from water to skin  is 6913.5 J.

3 0
2 years ago
Other questions:
  • Roseanne heated a solution in a beaker as part of a laboratory experiment on energy transfer. After a while, she noticed the liq
    5·1 answer
  • Which statement correctly describes the relationship between frequency and wavelength?
    14·2 answers
  • Which condition in a nebula would prevent nuclear fusion?
    14·2 answers
  • A 92-kg rugby player running at 7.5 m/s collides in midair with a 112-kg player moving in the opposite direction. After the coll
    7·1 answer
  • What is the change in length of a 1400. m steel, (12x10^-6)/(C0) , pipe for a temperature change of 250.0 degrees Celsius? Remem
    11·1 answer
  • A person travels distance πR along the circumference
    6·2 answers
  • A viscous liquid is sheared between two parallel disks of radius �, one of which rotates with angular speed Ω, while the other i
    14·1 answer
  • Find the truth table for the circuit shown. Explain the working principle for all the inputs, briefly. Explain why D1 is used in
    9·1 answer
  • Assertion(A):The distance moved by an object in unit time is called its speed. Reason (R):Faster vehicles have higher speeds. i)
    15·1 answer
  • A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antin
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!