As we know that
here we know that
now from above equation we have
so image will form on left side of lens at a distance of 15 cm
This image will be magnified and virtual image
Ray diagram is attached below here
Good work on solving part a).
b) may look complicated, but it's not too bad.
It says that the body is 25% efficient in converting fat to mechanical energy.
In other words, only 25% of the energy we get from our stored fat shows up
in the physical, mechanical moving around that we do. (The rest becomes
heat, which dissipates into the environment as we keep our bodies warm,
breathe hot air out,and perspire.)
You already know how much mechanical energy the climber needed to lift
himself to the top of the mountain... 2.4x10⁶ joules.
That's 25% of what he needs to convert in order to accomplish the climb.
He needs to pull 4 times as much energy out of fat.
-- Fat energy required = 4 x (2.4 x 10⁶) = 9.6 x 10⁶ joules.
-- Amount stored in 1kg of fat = 3.8 x 10⁷ joules
-- Portion of a kilogram he needs to use = (9.6 x 10⁶) / (3.8 x 10⁷)
Note:
That much of a kilogram weighs about 8.9 ounces ... which shows why it's so
hard to lose weight with physical exercise alone. It also helps you appreciate
that fat is much more efficient at storing energy than batteries are ... that one
kilogram of fat stores the amount of energy used by a 100-watt light bulb, to
burn for 105 hours (more than 4-1/2 days ! ! !)
W=ΔKE , W=-5000j
KEinitial=(1/2)mv² , KEfinal=0j
ΔKE=-(1/2)mv²
-5000=-(1/2)(100kg)v²
v=10 m/s
Charge on can A is positive.
Charge on can C is negative.
Punctuation and capitalization are very useful things to pay attention to and this question would be a lot easier to understand if you had actually used both capitalization and punctuation. If I'm understanding the question, you have 3 metal can that are insulated from the environment and initially touching each other in a straight line. Then a negatively charged balloon is brought near, but not touching one of the cans in that line of cans. While the balloon is near, the middle can is removed. Then you want to know the charge on the can that was nearest the balloon and the charge on the can that was furthermost from the balloon.
As the balloon is brought near to can a, the negative charge on the balloon repels some of the electrons from can a (like charges repel). Some of those electrons will flow to can b and in turn flow to can c. Basically you'll have a charge gradient that's most positive on that part of the can that's closest to the balloon, and most negative on the part of the cans that's furthest from the balloon. You then remove can B which causes cans A and C to be electrically isolated from each other and prevents the flow of elections to equalize the charges on cans A and C when the balloon is removed. So you're left with a deficiency of electrons on can A, so can A will have a positive overall charge, and an excess of electrons on can C, so can C will have a negative overall charge.
Answer:
a. y(x,t)= 2.05 mm cos[( 6.98 rad/m)x + (744 rad/s).
b. third harmonic
c. to calculate frequency , we compare with general wave equation
y(x,t)=Acos(kx+ωt)
from ωt=742t
ω=742
ω=2*pi*f
742/2*pi
f=118.09Hz
Explanation:
A fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)=2.30mmcos[(6.98rad/m)x+(742rad/s)t]. Being more practical-minded, you measure the rope to have a length of 1.35 m and a mass of 3.38 grams. Assume that the ends of the rope are held fixed and that there is both this traveling wave and the reflected wave traveling in the opposite direction.
A) What is the wavefunction y(x,t) for the standing wave that is produced?
B) In which harmonic is the standing wave oscillating?
C) What is the frequency of the fundamental oscillation?
a. y(x,t)= 2.05 mm cos[( 6.98 rad/m)x + (744 rad/s).
b. lambda=2L/n
when comparing the wave equation with the general wave equation , we get the wavelength to be
2*pi*x/lambda=6.98x
lambda=0.9m
we use the equation
lambda=2L/n
n=number of harmonics
L=length of string
0.9=2(1.35)/n
n=2.7/0.9
n=3
third harmonic
c. to calculate frequency , we compare with general wave equation
y(x,t)=Acos(kx+ωt)
from ωt=742t
ω=742
ω=2*pi*f
742/2*pi
f=118.09Hz
