Answer:
4 days
either multiply 128 by .5 until you get to 2 counting each time or use 2 formulas ln(n2/n1)=-k(t2-t1) to get k then input k into ln(2)=k*t1/2
n2 is final amount and n1 is beginning and t is either time elapsed as in the first formula or the actual half life that is t1/2
Explanation:
Answer:
The work required is -515,872.5 J
Explanation:
Work is defined in physics as the force that is applied to a body to move it from one point to another.
The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.
Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).
The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:


In this case:
- W=?
- m= 2,145 kg
- v2= 12

- v1= 25

Replacing:

W= -515,872.5 J
<u><em>The work required is -515,872.5 J</em></u>
The appropriate response is the third one. A generator is utilized to enact the control poles which contain radioactive isotopes. Once initiated, these isotopes start an atomic splitting chain response. Water in a cooling tank monitors the rate of response as electrons radiated from the response are encouraged through wires to homes and organizations.
Let say for every 5 s of time interval the speed will remain constant
so it is given as
v(mi/h) 16 21 23 26 33 30 28
now we have to convert the speed into ft/s as it is given that 1 mi/h = 5280/3600 ft/s
so here we will have
v(ft/s) 23.5 30.8 33.73 38.13 48.4 44 41.1
now for each interval of 5 s we will have to find the distance cover for above interval of time



so here it will cover 1298.1 ft distance in 30 s interval of time