No it's the quite opposite simple
Draw a circuit that contains 2 batteries, three lights in parallel and a switch that controls the whole circuit.
Answer:
The work done by the drag force is given by 29.96 J
Explanation:
Given :
Thrust force
N
Displacement
m
Mass of rocket
Kg
From work energy theorem,


Where
thrust work
gravitational work

After cutoff kinetic energy is converted into potential energy,

Put value of KE

Work done by drag force is given by,

J
Therefore, the work done by the drag force is given by 29.96 J
Answer:
Explanation:
I got everything but i. Don't know why but it's eluding me. So let's do everything but that.
a. PE = mgh so
PE = (2.5)(98)(14) and
PE = 340 J
b.
so
and
KE = 250 J
c. TE = KE + PE so
TE = 340 + 250 and
TE = 590 J
d. PE at 8.7 m:
PE = (2.5)(9.8)(8.7) and
PE = 210 J
e. The KE at the same height:
TE = KE + PE and
590 = KE + 210 so
KE = 380 J
f. The velocity at that height:
and
so
v = 17 m/s
g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:
590 = KE + PE and
PE = (2.5)(9.8)(11.6) so
PE = 280 then
590 = KE + 280 so
KE = 310 then
and
so
v = 16 m/s
h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:
and
26 = 0 + 9.8t and
26 = 9.8t so the time at 26 m/s is
t = 2.7 seconds. Now we use that in the equation for displacement:
Δx =
and filling in the time the object was at 26 m/s:
Δx = 0t +
so
Δx = 36 m
i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.
The rest energy of a particle is

where

is the rest mass of the particle and c is the speed of light.
The total energy of a relativistic particle is

where v is the speed of the particle.
We want the total energy of the particle to be twice its rest energy, so that

which means:


From which we find the ratio between the speed of the particle v and the speed of light c:

So, the particle should travel at 0.87c in order to have its total energy equal to twice its rest energy.