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SpyIntel [72]
3 years ago
7

I need help with these questions.

Physics
1 answer:
strojnjashka [21]3 years ago
4 0

I'll go ahead and answer the ones here without an answer. For reference, the half-life formula is <em>final amount = original amount(1/2)^(time/half-life)</em>

<em />

4) 12.5g

x = 100(1/2)^(63/21)

5) 50g

3.125 = x(1/2)^(0.1/0.025)

6) 500g

x = 4000(1/2)^(525/175)

7) 0.24g

0.06 = x(1/2)^(11430/5730)

8) 125g

x = 1000(1/2)^(17100/5700)

Hope this helps! :)

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A 50 kg stcel ball is hanging from two cables that make 120 degrees with each
FrozenT [24]

Answer:

T1 = 490.5 [N], T2 = 490.5[N]

Explanation:

First, we must draw a free body diagram of the steel ball hanging and the two wires holding it as well as the angle forming the wires between them.

The free-body diagram can be seen in the attached image.

As the cables are symmetrical with respect to the vertical axis, the force in cables 1 and 2 is equal, so when performing a force sum equal to zero on the Y-axis, we can find the force value of any cable.

The solution of the equations can be seen in the attached image

3 0
4 years ago
Why the object will not return to its original position?<br>why does the centre of mass fall?​
dusya [7]

Answer:

the object will not return to the original position because it will not have any forces helping it to go up the hill

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Hope this helps!! have a great day

Explanation:

8 0
3 years ago
A water bug is suspended on the surface of a pond by surface tension (water does not wet the legs). The bug has six leg, and eac
Crazy boy [7]

Answer:

m = 2.2 x 10⁻⁴ kg = 0.22 g

Explanation:

The surface tension of water is 0.072 N/m. So in order for the bug to avoid sinking, its weight per unit length of contact must be no more than the surface tension of water. Therefore,

Weight\ of bug\ per\ unit\ length = Surface\ Tension\ of\ Water\\\frac{mg}{L} = Surface\ Tension\ of Water\\m = \frac{(Surface\ Tension\ of\ Water)(L)}{g}

where,

m = mass of bug = ?

g = acceleration due to gravity = 9.81 m/s²

L = Contact length = (contact length of each leg)(No. of Legs) = (5 mm)(6)

L = 30 mm = 0.03 m

Therefore,

m = \frac{(0.072\ N/m)(0.03\ m)}{9.81\ m/s^{2}} \\

<u>m = 2.2 x 10⁻⁴ kg = 0.22 g</u>

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