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2 years ago

6

A sample from a meteorite that landed on Earth has been analyzed, and the result shows that out of every 1,000 nuclei of potassi

um-40 originally in the meteorite, only 250 are still present, meaning they have not yet decayed. How old is the meteorite (in yr)?

1 answer:

Lera25 [3.4K]2 years ago

5 0

I just did 1,000 nuclei divided by 250 and got 4.
The answer is 4 years.
Hope this helps!
Please give Brainliest!

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7-10/2<br><br> A.2<br> B.-13<br> C.12

vagabundo [1.1K]

=7-10/2

=7- 5

= 2

A. 2

Hope this helps

3 0

3 years ago

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A helium weather balloon is filled to a volume of 219 m3 on the ground, where the pressure is 754 torr and the temperature is 29

Novosadov [1.4K]

Answer:

$V_2 = 606.9 m^3$

Explanation:

By ideal gas equation law we know that

PV = nRT

now we know that when balloon rises to certain level then the number of moles will remains same

so we can say

$n_1 = n_2$

$\frac{P_1V_1}{RT_1} = \frac{P_2V_2}{RT_2}$

now plug in all data to find the final volume of the balloon

$\frac{754\times 219}{R(298)} = \frac{210 \times V_2}{R(230)}$

$V_2 = \frac{230\times 754 \times 219}{210 \times 298}$

$V_2 = 606.9 m^3$

8 0

3 years ago

What is the mass of metal if it has a density of 12.459 hg/cm^3 and displaces 28.7cm^3 of water

natima [27]

Answer:

357.6g

Explanation:

Given parameters:

Density = 12.459g/cm³

Volume of metal = 28.7cm³

Unknown:

Mass of metal = ?

Solution:

The density of a substance is its mass per unit volume.

To find the mass;

Mass of metal = density x volume

Now insert the parameters and solve;

Mass of metal = 12.459 x 28.7 = 357.6g

7 0

3 years ago

A cheetah can run at a maximum speed 97.8 km/h and a gazelle can run at a maximum speed of 78.2 km/h. If both animals are runnin

emmainna [20.7K]

(1)

Cheetah speed: $v_c = 97.8 km/h=27.2 m/s$

Its position at time t is given by

$S_c (t)= v_c t$ (1)

Gazelle speed: $v_g = 78.2 km/h=21.7 m/s$

the gazelle starts S0=96.8 m ahead, therefore its position at time t is given by

$S_g(t)=S_0 +v_g t$ (2)

The cheetah reaches the gazelle when $S_c=S_g$ . Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle:

$v_c t=S_0 + v_g t$

$(v_c -v_g t)=S_0$

$t=\frac{S_0}{v_c-v_t}=\frac{96.8 m}{27.2 m/s-21.7 m/s}=17.6 s$

(2) To solve the problem, we have to calculate the distance that the two animals can cover in t=7.5 s.

Cheetah: $S_c = v_c t =(27.2 m/s)(7.5 s)=204 m$

Gazelle: $S_g = v_g t =(21.7 m/s)(7.5 s)=162.8 m$

So, the gazelle should be ahead of the cheetah of at least

$d=S_c -S_g =204 m-162.8 m=41.2 m$

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3 years ago

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Which of the following is an example of resourcefulness?

tatiyna

A. Discovering a quick way to handle a new problem

5 0

3 years ago