First example: book, m= 0.75 kg, h=1.5 m, g= 9.8 m/s², it has only potential energy Ep,
Ep=m*g*h=0.75*9.8*1.5=11.025 J
Second example: brick, m=2.5 kg, v=10 m/s, h=4 m, it has potential energy Ep and kinetic energy Ek,
E=Ep+Ek=m*g*h + (1/2)*m*v²=98 J + 125 J= 223 J
Third example: ball, m=0.25 kg, v= 10 m/s, it has only kinetic energy Ek
Ek=(1/2)*m*v²=12.5 J.
Fourth example: stone, m=0.7 kg, h=7 m, it has only potential energy Ep,
Ep=m*g*h=0.7*9.8*7=48.02 J
The order of examples starting with the lowest energy:
1. book, 2. ball, 3. stone, 4. brick
Answer:
a. one-half as great
Explanation:
The power developed by the first lifter is one-half as great as that of the second person.
Power is defined as the rate at which work is done;
Power = 
Since the two lifters do the same work at different time, let us estimate their power;
P₁ = 
P₂ = 
We see that for P₁, power is half of the work done whereas in P₂ power is the same as the work done.
Therefore,
The power of the first weight lifter is one-half the second lifter.
Answer:
Landed before it explodes
Explanation:
vf = vi + at,
0 = 145 - (9.8)t,
t = 14.79 s (Time to reach highest point)
14.79 x 2 = 29.59 s (Time to land on the ground)
It will have landed before it explodes because both the time to reach the highest point and the time to land on the ground are less than 32 seconds.
Answer:
See below
Explanation:
<u>I will use 3 x 10^8 m/s for speed or wave</u>
speed = wavelength * frequency
3 x 10^8 = w * 7.34 x 10^2 <====== are you sure this isn't KILO Hz ?
w = <u>408719. 3 meters </u>
Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V
Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:
h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.
In order to calculate the cutoff wavelength we have to consider that Ek=0
in this case h*ν=W
(h*c)/λ=4.52 eV
λ= (h*c)/4.52 eV
λ= (1240 eV*nm)/(4.52 eV)=274.34 nm
From this h*ν = Ek+W; we can calculate the kinetic energy for a radiation wavelength of 198 nm
then we have
(h*c)/(λ)-W= Ek
Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV
Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this acts to slow down the ejected electrons from the catode.
