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mixas84 [53]
2 years ago
11

The half life of Po-218 is three minutes. How much of a 2.0 gram sample remains after 15 minutes? Suppose you wanted to buy some

of this isotope, and it required half an hour for it to reach you. How much should you order if you needed to use 0.10 gram of this material?
Physics
1 answer:
Greeley [361]2 years ago
5 0

After 15 minutes :

M = M₀ (1/2)^(t/T)

M = 2 (1/2)^(15/3)

M = 0.0625 gr

In order to get 0.1 gr :

M = M₀ (1/2)^(t/T)

0.1 = M₀ (1/2)^(30/3)

0.1 = Mo / 1024

Mo = 102.4 gr

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What is the momentum of a 0.5 kg ball which is thrown at 20 m/s?
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7 0
3 years ago
One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction
goblinko [34]

Answer:

The magnitud of the force is 124.8N.

Explanation:

First we have to find the value of the static friction coefficient, when the external force F is applied to upper block (i will call it A Block) we have a free body diagram as the one shown in the figure i attached, so since this block has no aceleration in any direction the force F should be equal to the friction force between A and B block, one we noticed this we can use the equation for the Friction force to find the coefficient:

0=F-FrictionAB

F=FrictionAB=Nab*μs

and again, since the block has no acceleration the normal between A and B block should be equal to the weigth of the first block, so we have:

0=Nab-W

Nab=W=mg

replacing this we have:

F=μs*Nab=μs*mg=41.6N

and  μs=41.6N/(mg)

now it's time to see the free body diagram for the b block, if we now apply the F force to the B block the diagram should look like in the figure.

the color of the arrow gives you an idea of where the force comes from, the blue ones comes from the B block, the red ones from the A block and the brown ones from the ground.

now for the B block you can see two friction forces, one for the ground and one for the A block, both of these directed bacwards, and two normal forces, again one for the ground and one for the A block but the normal force for the A block is aiming downwards.

again we use the fact that the block is not accelerating in any direction so the sum of the forces in x and y direction have to be 0, so:

F-Friction1(ground)-Friction2(AB)=0

This is the new external F force that we are looking for:

F=Friction1(ground)+Friction2(AB)

we know Friction2(AB) because we found that in the previous block so:

F=Friction1(ground)+mg*μs

for the other friction we have to use the equation:

Friction(ground)=N(ground)*μs

from y axis we have:

N(ground)-w-Normal(AB)=0

N(ground)=w+Normal(AB)

we found the value of Normal(AB) with the previous block so:

N(ground)=mg+mg=2mg

and:

Friction(ground)=2mg*μs

F=Friction(ground)+mg*μs

F=2mg*μs+μs*mg=3mg*μs

and since: μs*mg=41.6N

the new F force would be:

F=3mg*μs=41.6*3=124.8N

4 0
3 years ago
Can someone help? I’ll give brainlest .
MrRissso [65]

Answer:

I think it’s 6.8 m/s2

Explanation:

Please give me brainlist if the answer is right.

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3 years ago
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zloy xaker [14]

Option C i.e scattering is the correct answer.

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4 0
4 years ago
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Suppose that a particular artillery piece has a range R = 9880 yards . Find its range in miles. Use the facts that 1mile=5280ft
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Answer:

R = 9880 yd * 3 ft/yd / 5280 ft/mi = 5.61 mi

If you do it in steps

R = 9880 yd * 3 ft/yd = 29640 ft

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6 0
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