Answer:
a)
b)
Explanation:
Given:
mass of bullet, 
compression of the spring, 
force required for the given compression, 
(a)
We know
where:
a= acceleration
we have:
initial velocity,
Using the eq. of motion:
where:
v= final velocity after the separation of spring with the bullet.
(b)
Now, in vertical direction we take the above velocity as the initial velocity "u"
so,
∵At maximum height the final velocity will be zero
Using the equation of motion:
where:
h= height
g= acceleration due to gravity
is the height from the release position of the spring.
So, the height from the latched position be:
Answer:
The difference between ice and steam in Celsius (Centigrade) is 100 deg.
So the difference between and 4 cm and 24 cm of the thread corresponds to 100 deg C.
So 8 cm is 4 cm greater than the ice point
4 cm / 20 cm = 1/5 since the steam point and the ice point are 20 cm apart
Then 1/5 * 100 deg C = 20 deg C the requested temperature
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.
Answer:
1. B
2. B
3. D
4. A
Hope this was correct! A lot of the answers are already in the article itself and the wording is just different. I suggest now that for information retainment, you read the article again with the correct points in mind and see if you can spot the points where the answers are stated!
The working distance gets shorter as the magnification gets bigger. In order to focus, the high-power objective lens must be significantly nearer to the specimen than the low-power lens. Magnification is negatively correlated with working distance.
Magnification change The magnification of a specimen is increased by switching from low power to high power. The magnification of an image is determined by multiplying the magnification of the objective lens by the magnification of the ocular lens, or eyepiece.
The geometry of the optical system connects the magnifying power, or how much the thing being observed seems expanded, and the field of view, or the size of the object that can be seen.
To know more about working distance
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