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2 years ago

6

Plutonium-241 is an isotope of plutonium that is highly radioactive and is used in some nuclear reactors and many nuclear weapon

s. It has a half-life of 14.4 years. If 1000g of Pu-241 is placed in storage at a weapons plant, how much will still be remaining after 28.8 years?

2 answers:

choli [55]2 years ago

7 0

We use the radioactive decay equation for this problem which is expressed as:

An = Aoe^-kt

An is the remaining amount after time t, Ao is the initial amount and k is a constant.

First, we determine the k from the half life as follows:

An/Ao = 1/2 = e^-k(14.4)
k = 0.04814

Then, we can calculate An after 28.8 yr.

An = 1000 e^-0.04814(28.8)
An = 250 g

Andrew [12]2 years ago

6 0

Answer:

250g

Explanation:

i just took the test

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An object is placed perpendicular to the the principal axis of convex lens of focal length 8,the distance of the object from the

Digiron [165]

Answer:

v = 24 cm and inverted image

Explanation:

Given that,

The focal length of the object, f = +8 cm

Object distance, u = -12 cm

We need to find the position &nature of the image. Let v be the image distance. Using lens formula to find it :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Put all the values,

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{8}+\dfrac{1}{(-12)}\\\\v=24\ cm$

So, the image distance from the lens is 24 cm.

Magnification,

$m=\dfrac{v}{u}\\\\m=\dfrac{24}{-12}\\\\m=-2$

The negative sign of magnification shows that the formed image is inverted.

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2 years ago

Masses A and B rest on very light pistons that enclose a fluid.There is no friction between the pistons and the cylinders they f

RSB [31]

Answer:

D)Not enough information

Explanation:

According to Pascal's principle, the pressure exerted on the two pistons is equal:

$p_A = p_B$

Pressure is given by the ratio between force F and area A, so we can write

$\frac{F_A}{A_A}=\frac{F_B}{A_B}$

The force exerted on each piston is just equal to the weight of the corresponding mass: $F=W=mg$ , where m is the mass and g is the gravitational acceleration. So the equation becomes

$\frac{m_A g}{A_A}=\frac{m_B g}{A_B}$

Now we can rewrite the mass as the product of volume, V, times density, d:

$\frac{V_A d_A g}{A_A}=\frac{V_B d_B g}{A_B}$

We also know that $A_B = 2.0 m^2\\A_A = 1.0 m^2$

So we can further re-arrange the equation (and simplify g as well):

$\frac{V_A d_A}{1}=\frac{V_B d_B}{2}$

$\frac{d_A}{d_B}=\frac{V_B}{2V_A}$

We are also told that block B has bigger volume than block A: $V_B > V_A$ . However, this information is not enough to allow us to say if the fraction on the right is greater than 1 or smaller than 1: therefore, we cannot conclude anything about the densities of the two objects.

3 0

2 years ago

Answer it pls!!!!!!!!!!!

Archy [21]

Answer:

Fractional error = 0.17

Percent error = 17%

F = 112 ± 19 N

Explanation:

Plug in the values to find the force:

F = (3.5 kg) (20 m/s)² / (12.5 m) = 112 N

Find the fractional error:

ΔF/F = Δm/m + 2Δv/v + Δr/r

ΔF/F = 0.1/3.5 + 2(1/20) + 0.5/12.5

ΔF/F = 0.17

Multiply by 100% to find the percent error:

ΔF/F × 100% = 17%

Solve for the absolute error:

ΔF = 0.17 × 112 N = 19 N

Therefore, the force is:

F = 112 ± 19 N

8 0

2 years ago

Why is velocity and not just speed important to these pilots

Alborosie

Velocity is a vector. Therefore, it depends on the direction. Pilots need to know the direction of wind, not just the speed. If the pilot is going South, and there's 5 mph wind going South, they'll be happy, but if the wind is going 5 mph North, they'll be going against the wind.

7 0

3 years ago

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Pie

Answer:

Part a)

$v = 7407.1 m/s$

Part b)

$v_{rel} = 1.05 \times 10^4 m/s$

Explanation:

Part a)

As we know that orbital velocity at certain height from the surface of Earth is given as

$v = \sqrt{\frac{GM}{R+h}}$

here we know that

$M = 5.98 \times 10^{24} kg$

$R = 6.37 \times 10^6 m$

$h = 900 km = 9.0 \times 10^5 m$

now we have

$v = \sqrt{\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{6.37 \times 10^6 + 9.0 \times 10^5}}$

$v = 7407.1 m/s$

Part b)

When a loose rivet is moving in same orbit but at 90 degree with the previous orbit path then in that case the relative speed of the rivet with respect to the satellite is given as

$v_{rel} = \sqrt{2} v$

$v_{rel} = 1.05 \times 10^4 m/s$

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2 years ago