The sample appears to have gone through 3 half-lives
1st half life: 1000 to 500 g
2nd half life: 500 to 250 g
3rd half life: 250 to 125 g
The duration of a half-life, therefore, can be inferred to be 66 ÷ (3) = 22 days.
After a 4th half life, there will be 125÷2= 62.5 g.
At this point, an additional 22 days will have passed, for a total of 88 days.
Answer is C.
Refer to the diagram shown below.
For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)
For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)
Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N
Answer:
T₂ = 339.06 N
T₃ = 276.57 N
Answer:
Buoyancy force and surface tension are the reactions that take places between soap and pepper experiment.
Explanation:
Surface tension:
The surface tension of a liquid is the tendency of liquid surfaces to resist an external force due to the cohesive nature of its molecules.
The pepper and soap experiment helps you to understand buoyancy force and surface tension.
Reaction between the pepper and soap is as following.
- The pepper flakes float because of buoyancy force. It makes the pepper flakes to move away to the edge of the plate.
- This happens because the liquid dish soap changes the surface tension of water.
- And The pepper flakes are so light, it floats on the water surface due to surface tension.
- when we add soap, it breaks the surface tension of water, but the water resists it. So they pull away from the soap along with the pepper flakes.
- This pushes the pepper away from your soap covered finger.
This is the reaction that take places between soap and pepper experiment.
Learn more about Pepper and soap experiment here:
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Explanation:
It is given that,
A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.
Mass of the Sun, 
Radius of Mercury's orbit, 
Radius of discovered planet, 

Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :




T = 4135214.625 s
or
T = 47.86 days
So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.
Answer:
Friction: is used to hang an object on the wall
Force: is what will determine if the object stays in place or not
Explanation: