You are given the half life of an isoptope which is 5.2yr and the numer of days that it will decay which is 30 months. You are required to find the initial mass of the isotope.
An = A0*exptaveln2/t1/2
30 = A0 exp^(30/12 * ln2/5.2yr)
A0 = 21.5grams
Answer:
The magnitude of the vector A is <u>51 m.</u>
Explanation:
Given:
The horizontal component of a vector A is given as:
The vertical component of a vector A is given as:
Now, we know that, a vector A can be resolved into two mutually perpendicular components; one along the x axis and the other along the y axis. The magnitude of the vector A can be written as the square root of the sum of the squares of each component.
Therefore, the magnitude of vector A is given as:
Now, plug in 44.4 for , 25.1 for and solve for the magnitude of A. This gives,
Therefore, the magnitude of the vector A is 51 m.
Answer:
d
Explanation:
I looked it up for you to help you out and it said it was d
Answer:
4.8 m/s
Explanation:
When she catches the train,
- They will have travelled the same distance.and
- Their speeds will be equal
The formula for the distance covered by the train is
d = ½at² = ½ × 0.40t² = 0.20t²
The passenger starts running at a constant speed 6 s later, so her formula is
d = v(t - 6.0)
The passenger and the train will have covered the same distance when she has caught it, so
(1) 0.20t² = v(t - 6.0)
The speed of the train is
v = at = 0.40t
The speed of the passenger is v.
(2) 0.40t = v
Substitute (2) into (1)
0.20t² = 0.40t(t - 6.0) = 0.40t² - 2.4 t
Subtract 0.20t² from each side
0.20t² - 2.4t = 0
Factor the quadratic
t(0.20t - 2.4) = 0
Apply the zero-product rule
t =0 0.20t - 2.4 = 0
0.20t = 2.4
(3) t = 12
We reject t = 0 s.
Substitute (3) into (2)
0.40 × 12 = v
v = 4.8 m/s
The slowest constant speed at which she can run and catch the train is 4.8 m/s.
A plot of distance vs time shows that she will catch the train 6 s after starting. Both she and the train will have travelled 28.8 m. Her average speed is 28.8 m/6 s = 4.8 m/s.