All categories

1 year ago

A radioactive source has a half life of 80s.

1 answer:

6 0

Lets make the original number of nuclides at the start is 100.

If 7/8 of 100 is decayed, that means 87.5 decayed.

$\frac{7}{8} \times 100 = 87.5$

And there is 1/8 left of the number of nuclide 100. Which is 12.5

$100 - 87.5 =12.5$

$\frac{1}{8} \times 100 = 12.5$

How many Half lifes passed for 100 to become 12.5 is 3 Half-Lives.

$100 \div 2 \div 2 \div 2 = 12.5$

Each Half-Life is 80 seconds so there is 240 seconds

$3 \times 80 = 240$ The answer is that it takes 240 seconds.

You might be interested in

A nonrelativistic electron is accelerated from rest through a potential difference. After acceleration the electron has a de Bro

aleksley [76]

Answer:

Potential difference though which the electron was accelerated is $2.67\times 10^{-6}\ uV\ .$

Explanation:

Given :

De Broglie wavelength , $\lambda=750\ nm.$

Plank's constant , $h=6.626\times 10^{-34}\ J.s \ .$

Charge of electron , $e=-1.6\times 10^{-19}\ C.$

Mass of electron , m=9.11\times 10^{-31}\ kg. $m=9.11\times 10^{-31}\ kg.$

We know , according to de broglie equation :

$\lambda=\dfrac{h}{mv}\\\\ v=\dfrac{h}{m\lambda}\\\\v=\dfrac{6.626\times 10^{-34}\ J.s \ }{9.11\times 10^{-31}\ kg\times 750\times 10^{-9}\ m }= 969.78\ m/s .$

Now , we know potential energy applied on electron will be equal to its kinetic energy .

Therefore ,

$qV=\dfrac{mv^2}{2}\\\\ V=\dfrac{mv^2}{2q}$

Putting all values in above equation we get ,

$V=2.67\times 10^{-6}\ uV .$

Hence , this is the required solution.

5 0

1 year ago

10. A girl pulls a wagon along a level path for a distance of 44 m. The handle of

Rina8888 [55]

The work done on the wagon is 3549 J

Explanation:

The work done by a force when moving an object is given by

$W=Fd cos \theta$

where :

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and of the displacement

In this problem we have the following data:

F = 87 N is the magnitude of the force

d = 44 m is the displacement of the wagon

$\theta=22^{\circ}$ is the angle between the direction of the force and the displacement

Substituting, we find the work done

$W=(87)(44)(cos 22^{\circ})=3549 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0

1 year ago

José is pinned against the walls of the Rotor, a ride with a radius of 3.00 meters that spins so fast that the floor can be remo

zaharov [31]

r = radius of the circle of the ride = 3.00 meters

v = linear speed of the person during the ride = 17.0 m/s

m = mass of the person in angular motion in the ride

L = angular momentum of the person in the ride = 3570 kg m²/s

Angular momentum is given as

L = m v r

inserting the values

3570 kg m²/s = m (17 m/s) (3.00 m)

m = 3570 kg m²/s/(51 m²/s)

m = 7 kg

hence the mass comes out to be 7 kg

8 0

1 year ago

The planet Neptune orbits the Sun. Its orbital radius is 30.130.130, point, 1 astronomical units (\text{AU})(AU)left parenthesis

lord [1]

Answer:

The distance the planet Neptune travels in a single orbit around the Sun is 60.2π AU.

Explanation:

As it is given that the Neptune's orbit is circular, the formula that we have to use is the circumference of a circle in order to find the distance it travels in a single orbit around the Sun. In other words, you can say that the circumference of the circle is equivalent to the distance it travels around the Sun in a single orbit.

The circumference of the circle = Distance Travelled (in a single orbit) = 2*π*R ---- (A)

Where,

R = Orbital radius (in this case) = 30.1 AU

Plug the value of R in the equation (A):

(A) => The circumference of the circle = 2*π*(30.1)

The circumference of the circle = 60.2π

Therefore, the distance the planet Neptune travels in a single orbit around the Sun is 60.2π AU.

5 0

1 year ago

A solid sphere of brass (bulk modulus of 14.0 ✕ 1010 N/m2) with a diameter of 2.20 m is thrown into the ocean. By how much does

astra-53 [7]

Answer: