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LenaWriter [7]
1 year ago
11

A radioactive isotope has a half-life of 2 hours. If a sample of the element contains 600,000 radioactive nuclei at 12 noon, how

many radioactive nuclei would be left at 6 pm?
Physics
1 answer:
storchak [24]1 year ago
7 0

Answer: There will be 75258 nuclei left at 6 pm.

Explanation:

a) half-life of the radioactive substance:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.69}{k}

k=\frac{0.69}{t_{\frac{1}{2}}}=\frac{0.693}{2hours}=0.346hours^{-1}

b) Expression for rate law for first order kinetics is given by:

A=A_0e^{-kt}

where,

k = rate constant  

t = time for decomposition = 6 hours ( from 12 noon to 6 pm)

A = activity at time t = ?

A_0 = initial activity  = 600, 000

A=600000\times e^{-0.346\times 6}

A=75258

Thus there will be 75258 nuclei left at 6 pm.

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When you are standing on Earth, orbiting the Sun, and looking at a broken cell phone on the ground, there are gravitational pull
Mandarinka [93]

Answer:

The answer is "Option b, c, and a".

Explanation:

Here that the earth pulls on the phone, as it will accelerate towards Earth when we drop it.

We now understand the effects of gravity:

F \propto  M\\\\F\propto  \frac{1}{r^2}\\\\or\\\\F \propto  \frac{M}{r^2}\\\\Sun (\frac{M}{r^2}) = \frac{10^{28}}{(10^9)^2} = 10^{10}

The force of the sun is, therefore, 10^{10} times greater and the proper sequence, therefore, option steps are:

b. Pull-on phone from earth

c. Pull-on phone from sun

a. Pull phone from you

5 0
1 year ago
A roller coaster car drops a maximum vertical distance of 35.4 m. Determine the maximum speed of the car at the bottom of that d
marissa [1.9K]

Answer:

The maximum speed of the car at the bottom of that drop is 26.34 m/s.

Explanation:

Given that,

The maximum vertical distance covered by the roller coaster, h = 35.4 m

We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :

mgh=\dfrac{1}{2}mv^2

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 35.4}

v = 26.34 m/s

So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.

8 0
2 years ago
A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences
NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, q=3\times 10^{-6}\ C

Mass of charge particle, m=2\times 10^{-6}\ C

Speed, v=5\times 10^{6}\ m/s

Acceleration, a=3\times 10^{4}\ m/s^2

We need to find the minimum magnetic field that would produce such an acceleration. So,

ma=qvB\ sin\theta

For minimum magnetic field,

ma=qvB

B=\dfrac{ma}{qv}

B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0
2 years ago
A worker pushes a 7 kg shipping box along a roller track. Assume friction is small enough to be ignored because of the rollers.
Bingel [31]

Answer:

a) Fₓ = 23.5 N

b) Net force = Fₓ

Explanation:

An image of the question as described is attached to this solution.

From the image attached, the forces acting on the box include the weight of the box, the normal reaction of the surface on the box, the applied force on the box and the Frictional force opposing the motion of the box (which is negligible and equal to 0)

a) From the diagram, the horizontal component of the force is

Fₓ = 25 cos 20° = 23.49 N = 25 N

b) Again, from the diagram attached, doing a force balance on the box, in the horizontal direction, we obtain

Net force = Fₓ - Frictional force

But frictional force is 0 N

Net force = Fₓ

Hope this Helps!!!

6 0
2 years ago
A solenoid of length 18 cm consists of closely spaced coils of wire wrapped tightly around a wooden core. The magnetic field str
Kisachek [45]

Answer:

B_2 = 1.71 mT

Explanation:

As we know that the magnetic field near the center of solenoid is given as

B = \frac{\mu_0 N i}{L}

now we know that initially the length of the solenoid is L = 18 cm and N number of turns are wounded on it

So the magnetic field at the center of the solenoid is 2 mT

now we pulled the coils apart and the length of solenoid is increased as L = 21 cm

so we have

\frac{B_1}{B_2} = \frac{L_2}{L_1}

now plug in all values in it

\frac{2.0 mT}{B_2} = \frac{21}{18}

B_2 = 1.71 mT

3 0
1 year ago
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