Answer:
I think answer is zero
bcz momentum=mass×velocity
body was initially at rest it means its velocity is zero
30×0=0
Answer:
t = 5.59x10⁴ y
Explanation:
To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:
(1)
<em>where 
: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>
To find A₀ we can use the following equation:
(2)
<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>
From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:
<em>where 
: is the tree's carbon mass, 
: is the Avogadro's number and 
: is the ¹²C mass. </em>
Similarly, from equation (2) λ is:
<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>
So, the initial activity A₀ is:
Finally, we can calculate the time from equation (1):
I hope it helps you!
Answer:
3430000 J
Explanation:
The formula for potential energy is PE=mgh.
M being the mass, g being the force of gravity, and h being the height.
First thing you want to do is convert 250 kg to g (grams).
From there you get 25000g and you have to multiply that by 14m and 9.8m/s^2 (the force of gravity is constant, at least on earth).
Answer:
It can only display one record at a time
Explanation:
Form ;
1. This is a document with spaces (also called placeholders or fields ) in which a series of documents with similar content can be written or selected.
2.This is the most popular method of data entry
3.It may contain images in the background.
4.This can be sorted data regardless of its source of information.
Only option C is wrong.
Therefore the answer C is correct.
Answer:
21 m/s.
Explanation:
The computation of the wind velocity is shown below:
But before that, we need to find out the angles between the vectors
53° - 35° = 18°
Now we have to sqaure it i.e given below
v^2 = 55^2 + 40^2 - 2 · 55 · 40 · cos 18°
v^2 = 3025 + 1600 - 2 · 55 · 40 · 0.951
v^2 = 440.6
v = √440.6
v = 20.99
≈ 21 m/s
Hence, The wind velocity is 21 m/s.
