Plutonium has a half life of 2.4 × 104 years. How long does it take for 99.0% of the plutonium to decay?

pashok25 [27]

As far as I'm concerned, this is a bogus question, or at least a severely corrupted one.

The three numbers given can NOT all be true on Earth.

-- It rolled off the table at 7.6 m/s . By golly, there you are! Its initial horizontal velocity is 7.6 m/s, and it has no vertical velocity until it leaves the table.

-- There are no horizontal forces that we're aware of acting on the object. So it maintains the same horizontal velocity for the rest of the story. It's 10.5m away from the table in (10.5 m) / (7.8 m/s) = 1.35 second .

-- Vertically, it's just an object dropped from 17.6m off the floor. Shockingly, the distance it falls in time 'T' is (1/2 g) T². In 1.35 second, that's 8.88 meters ! . . . only about halfway to the floor !

-- In order to fall 17.6 m to the floor, it would need 1.89 seconds. In that length of time, however, it would travel (7.8 m/s) x (1.89 s) = 14.78 m away from the base of the table.

So you see, either . . .

-- the table is NOT 17.6m tall, or

-- the object does NOT roll off of the table at 7.8 m/s, or

-- it does NOT land 10.5 m away from the base of the table.

OR . . .

-- the table is not on Earth, and gravity is not 9.8 m/s² !

We often see questions posted on Brainly with not enough given information, OR with some information given that's not needed because it's not involved the answer.

THIS one is different, and it's unusual. In this one, we have too much given information, we can't ignore any of it because it's all related, but it's inconsistent and it CAN't all be true.

(Unless the whole story takes place on a mystery planet that is not Earth. Which I'm not going to take the time and effort right now to figure out what the acceleration of gravity has to be in order to make all of the given information compatible.)

7 0

3 years ago

Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit

Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

$= 822486 \times 10^{-8}$

$=0.822 \times 10^{-2} \ km/s$

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

$= 153.84 \times 2371 \times 10^{-8}$

$=0.364 \times 10^{-2} \ km/s$

= 3.64 m/s

3 0

3 years ago

Which scientific activities will Juno conduct on its trip to Jupiter? Check all that apply.

Arisa [49]

Explanation:

Hope this helps,

Juno entered a polar orbit of Jupiter on July 5th 2016 UTC, to begin a scientific investigation of the planet. After completing its mission, Juno will be intentionally deorbited into Jupiters atmosphere. Junos mission is to measure Jupiters composition, gravitational field, magnetic field, and polar magnetosphere.

5 0

3 years ago

Read 2 more answers

A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0o above the horizontal towar

Artist 52 [7]

Answer:

a) v₀ = 32.64 m / s , b) x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

x = v₀ₓ t

y = $v_{oy}$ t - ½ g t²

We look for the components of speed with trigonometry

sin 43 = v_{oy} / v₀

cos 43 = v₀ₓ / v₀

v_{oy} = v₀ sin 43

v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

t = x / v₀ cos 43

y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

y = x tan 43 - ½ g x² / v₀² cos² 43

1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

v₀ = √ (35280 / 33.11)

v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

y = $v_{oy}$ t - ½ g t²

.y = v₀ sin43 t - ½ g t²

25 = 32.64 sin 43 t - ½ 9.8 t²

4.9 t² - 22.26 t + 25 = 0

t² - 4.54 t + 5.10 = 0

We solve the second degree equation

t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

t = (4.54 ± 0.46) / 2

t₁ = 2.50 s

t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

x = v₀ₓ t

x = v₀ cos 43 t

x = 32.64 cos 43 2.50

x = 59.68 m

8 0

4 years ago

The mechanical advantage to simple machines is that they allow a decreased input force to create a larger output force. please s

Rina8888 [55]

The mechanical advantage to simple machines is that they allow a decreased input force to create a larger output force.

TRUE

5 0

3 years ago