Answer:
5730 years
Explanation:
The half life of carbon-14 is 5730 years. If 50% of the carbon-14 remains, then exactly 1 half life has passed.
The half-life equation is:
A = A₀ (½)^(t / T)
where A is the remaining amount,
A₀ is the initial amount,
t is time,
and T is the half life.
In this case, A = ½ A₀ and T = 5730.
½ A₀ = A₀ (½)^(t / 5730)
½ = (½)^(t / 5730)
1 = t / 5730
t = 5730
Answer:
1065 Kgm-3
Explanation:
We can determine the relative density of the athlete from the formula;
Relative density of athlete = weight of athlete in air/upthrust on athlete
Since weight of athlete in air= 690 N
Weight of athlete in water = 42 N
Upthrust on athlete= weight in air - weight in water
Upthrust on athlete= 690 N - 42 N = 648 N
Relative density of athlete= 690 N / 648 N
Relative density of athlete= 1.065
Therefore, average density of the athlete= relative density × density of water = 1.065 × 1000 Kgm-3 = 1065 Kgm-3
We have: K.E. = mv² / 2
Here, m = 1500 Kg
v = 2 m/s
Substitute their values in the formula,
K.E. = 1500 ×2² / 2
K.E. = 6000 / 2
K.E. = 3000 J or 3 KJ
Finally, answer of your question would be 3000 Joule or 3 Kilojoule
Hope this helps!
Light travels<span> as a </span>wave<span>. But unlike sound </span>waves<span> or water </span>waves<span>, it does not need any matter or material to carry its energy along. This means that </span>light<span> can </span>travel<span> through a vacuum—a completely airless space.</span>
Answer:
Explanation:
Conservation of momentum during the collision
0.023(230) + 2.0(0.0) = 0.023(170) + 2.0v
v = 0.69 m/s
The initial block kinetic energy will be converted to friction work
½mv² = Fd = μmgd
½(2.0)(0.69²) = 0.15(2.0)(9.8)d
d = 0.1619387... m
d = 16 cm