Answer:
22,800 years
Explanation:
Half life equation:
A = A₀ (½)^(t / T)
where A is the final amount,
A₀ is the initial amount,
t is time,
and T is the half life.
0.0625 = (½)^(t / 5700)
log 0.0625 = (t / 5700) log 0.5
4 = t / 5700
t = 22,800
It takes 22,800 years.
Answer:
When friction hits the floor, the oil bounces back up but then down. Then the oil spreads apart. Like an earthquake. Not really though. And try not to spill oil for the first place.
Hope I helped! Brainiest plz! Hope you make an 100% and have a wonderful day! -Amelia♥
Answer:
C. 30.6m
Explanation:
To find the height of the tower, we are to use Newtons law of motion to solve this problem. Since the penny is falling from the top of the tower, it is acted by the acceleration due to gravity. The formula to be used is:

Where H is the height of the tower, t is the time taken to hit the ground, u is the initial velocity and g is the acceleration due to gravity.
Given that, t = 2.5 s, g =9.8 m/s², u = 0 m/s (at the top of tower)

The answer is (B. The study of Matter and Energy) but technically you could consider physics all of these as engineering is based on physics and that would be the study of inventions, chemistry and biology were both discovered because of physics, and physics invokes more math than any other subject as it applies math to the entire Universe.
R is proportional to the length of the wire:
R ∝ length
R is also proportional to the inverse square of the diameter:
R ∝ 1/diameter²
The resistance of a wire 2700ft long with a diameter of 0.26in is 9850Ω. Now let's change the shape of the wire, adding and subtracting material as we go along, such that the wire is now 2800ft and has a diameter of 0.1in.
Calculate the scale factor due to the changed length:
k₁ = 2800/2700 = 1.037
Scale factor due to changed diameter:
k₂ = 1/(0.1/0.26)² = 6.76
Multiply the original resistance by these factors to get the new resistance:
R = R₀k₁k₂
R₀ = 9850Ω, k₁ = 1.037, k₂ = 6.76
R = 9850(1.037)(6.76)
R = 69049.682Ω
Round to the nearest hundredth:
R = 69049.68Ω