If no frictional work is considered, then the energy of the system (the driver at all positions is conserved.
Let
position 1 = initial height of the diver (h₁), together with the initial velocity (v₁).
position 2 = final height of the diver (h₂) and the final velocity (v₂).
The initial PE = mgh₁ and the initial KE = (1/2)mv₁²
where g = acceleration due to gravity,
m = mass of the diver.
Similarly, the final PE and KE are respectively mgh₂ and (1/2)mv₂².
PE in position 1 is converted into KE due to the loss in height from position 1 to position 2.
Therefore
(KE + PE) ₁ = (KE + PE)₂
Evaluate the given answers.
A) The total mechanical energy of the system increases.
FALSE
B) Potential energy can be converted into kinetic energy but not vice versa.
TRUE
C) (KE + PE)beginning = (KE + PE) end.
TRUE
D) All of the above.
FALSE
Answer:
0.8 seconds
Explanation:
F=ma
Let x be the seconds the force is applied.
m = 20kg
F = 50 Newtons (kg*m/sec^2)
acceleration, a, is provided for x seconds to increase the speed from 1 m/s to 3 m/s, an increase of 2m/s
Let's calculate the acceleration of the cart:
F=ma
(50 kg*m/s^2) = (20kg)*a
a = 2.5 m/s^2
---
The acceleration is 2.5 m/s^2. The cart increases speed by 2.5 m/s every second.
We want the number of seconds it takes to add 2.0 m/sec to the speed:
(2.5 m/s^2)*x = 2.0 m/s
x = (2.0/2.5) sec
x = 0.8 seconds
It is b! the lithosphere is the crust and upper mantle
The main formula is given by Eb/nucleon = Eb/ mass of nucleid
as for <span>52He, the mass is 5
so by applying Einstein's formula Eb=DmC², Eb=</span><span>binding energy
</span><span>52He-----------> 2 x 11p + 3 x10n is the equation bilan
</span>so Dm=2 mp + (5-2)mn-mnucleus, mp=mass of proton=1.67 10^-27 kg
mn=mass of neutron=<span>1.67 10^-27 kg
</span><span>m nucleus= 5
Dm= 2x</span>1.67 10^-27 kg+ 3x<span>1.67 10^-27 kg-5= - 4.9 J
Eb= </span> - <span>4.9 J x c²= -4.9 x 9 .10^16= - 45 10^16 J
so the answer is Eb /nucleon = Eb/5= -9.10^16 J, but 1eV=1.6 . 10^-19 J
so </span><span>-9.10^16 J/ 1.6 10^-19= -5.625 10^35 eV
the final answer is </span><span>Eb /nucleon </span><span>= -5.625 x10^35 eV</span>