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2 years ago

If you have a 200g sample of a radioactive isotope that has a half life of 50 years, hc

Physics

1 answer:

Marizza181 [45]2 years ago

8 0

Answer:

The correct option is;

a. 12.5 g

Explanation:

The given parameters are;

The initial mass of the radioactive substance = 200 g

The half life of the radioactive substance = 50 years

The time duration for the disintegration = 200 years

The formula for half life is given as follows;

$N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}}$

Where:

N(t) = Quantity of the remaining substance

N₀ = Initial radioactive substance quantity = 200 g

t = Time duration = 200 years

$t_{1/2}$ = Half life of the radioactive substance 50 years

Therefore, we have;

$N(t) =200 \times \left (\dfrac{1}{2} \right )^{\dfrac{200}{50} } =200 \times \left (\dfrac{1}{2} \right )^4 = 12.5 \ g$

Therefore, at the end of 200 years, the quantity left = 12.5 g.

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A stone is dropped into a river from a bridge 41.7 m above the water. Another stone is thrown vertically down 1.80 s after the f

hram777 [196]

Answer:

31.75 m/s

Explanation:

h = 41.7 m

Let the initial velocity of the second stone is u

Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.

For first stone:

Use second equation of motion

$h=ut+\frac{1}{2}gt^2$

Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7

So, 41.7= 0 + 0.5 x 9.8 x t^2

41.7 = 4.9 t^2

t = 2.92 s ..... (1)

For second stone:

Use second equation of motion

$h=ut+\frac{1}{2}gt^2$

Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity

$h=u\left ( t-1.8 \right )+4.9\left ( t-1.8 \right )^2$ .... (2)

By equation the equation (1) and (2), we get

$41.7=1.12 u +4.9 \times 1.12^{2}$

u = 31.75 m/s

5 0

2 years ago

If a quasar is moving away from us at v/c = 0.8, what is the measured redshift?

Delicious77 [7]

Answer:

The measured redshift is z =2

Explanation:

Since the object is traveling near light speed, since v/c = 0.8, then we have to use a redshift formula for relativistic speeds.

$z= \sqrt{\cfrac{c+v}{c-v}}-1$

Finding the redshift.

We can prepare the formula by dividing by lightspeed inside the square root to both numerator and denominator to get

$z= \sqrt{\cfrac{1+\cfrac vc}{1-\cfrac vc}}-1$

Replacing the given information

$z= \sqrt{\cfrac{1+0.8}{1-0.8}}-1$

$z= \sqrt{\cfrac{1.8}{0.2}}-1\\z= \sqrt{9}}-1\\z=3-1\\z=2$

Thus the measured redshift is z = 2.

4 0

2 years ago

A 15-watt bulb is connected to a circuit that has a total of 60. Ω of resistance. How many electrons are passing through that bu

Mariulka [41]

Answer:

3.2075*10^16

Explanation:

Q=P/V just search up a converter and youll get 30V and so you do 15/30 which is a half and a single coulomb is 6.415*10^16 so you half it. I belive this is correct if you dont belive me wait for someone else smarter to answer and compare.

3 0

1 year ago

A hot cube of iron was heated up using 1500 J of thermal energy and was placed in a beaker of water. Before it was heated, the i

Mariana [72]

Answer:

451.13 J/kg.°C

Explanation:

Applying,

Q = cm(t₂-t₁)............... Equation 1

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

Substitute these values into equation 2

c = 1500/[0.133(45-20)]

c = 1500/(0.133×25)

c = 1500/3.325

c = 451.13 J/kg.°C

4 0

2 years ago

A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 5100 MJ with respect to Eart

serious [3.7K]

To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,

$E_{initial} = E_{final}$

$PE_{initial}+KE_{initial} = PE_{final}+KE_{final}$

Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that

$KE_{final} = (PE_{initial}+KE_{initial} )-PE_{final}$

$KE_{final} = (5100+4200)-5700$

$KE_{final} = 3600MJ$

Therefore the final kinetic energy is 3600MJ

5 0

2 years ago