Answer:
The half-life is 
Explanation:
Using the decay equation we have:

Where:
- λ is the decay constant
- A(0) the initial activity
- A is the activity at time t
We know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that


Taking the natural logarithm on each side we have:


Now, the relationship between the decay constant λ and the half-life t(1/2) is:




I hope it helps you!
Answer:
a) t = 2.0 s, b) x_f = - 24.56 m, Δx = 16.56 m
Explanation:
This is an exercise in kinematics, the relationship of position and time is indicated
x = 5 t³ - 9t² -24 t - 8
a) ask when the velocity is zero
speed is defined by
v =
let's perform the derivative
v = 15 t² - 18t - 24
0 = 15 t² - 18t - 24
let's solve the quadratic equation
t =
t1 = -0.8 s
t2 = 2.0 s
the time has to be positive therefore the correct answer is t = 2.0 s
b) the position and distance traveled for a = 0
acceleration is defined by
a = dv / dt
a = 30 t - 18
a = 0
30 t = 18
t = 18/30
t = 0.6 s
we substitute this time in the expression of the position
x = 5 0.6³ - 9 0.6² - 24 0.6 - 8
x = 1.08 - 3.24 - 14.4 - 8
x = -24.56 m
we see that all the movement is in one dimension so the distance traveled is the change in position between t = 0 and t = 0.6 s
the position for t = 0
x₀ = -8 m
the position for t = 0.6 s
x_f = - 24.56 m
the distance
ΔX = x_f - x₀
Δx = | -24.56 -(-8) |
Δx = 16.56 m
A because its about an opening in the sea floor and tectonic plates are the only thing.
Answer:
4m East
Explanation:
26-29= -3 (so that means 3m west)
-3+7= 4 (so 4m East)
hope it helps
Answer: (b)
Explanation:
Given
mass of Student M=40 kg
mass of box m=3 kg
velocity of box v=8 m/s
suppose u is the velocity of the student after the throw
As there is no external force is applied so momentum is conserved
Initial momentum=final momentum=0
0=Mu+mv
Mu=-mv

Here negative sign indicates the velocity is opposite in direction i.e. opposite to the direction of the box