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596277

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Posted: 13 years ago

#521

^ thanks so much I'll PM you quest. 3 and 4 which I was unsure about.

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596277

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Posted: 13 years ago

#522

I need help with Biology, about the microscope.

http://imageshack.us/photo/my-images/828/scan0001im.jpg/

I'll write the answers I've got in a few minutes.

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596277

Suspended

Posted: 13 years ago

#523

I've already done the diagram. Questions 2-14 If you could please go through and add anything I've missed or correct something that I've done wrong, that would be really appreciated.

2. a) Ocular Lens

b) Objective lens

c) Revolving nosepiece

d) Diaphragm

e) Stage

f) Condenser

g) Arm and Base

h) Stage clips

I)?

J) Fine adjustment Knob

3. Compound light microscope, Transmission electron microscope, Scanning electron microscope, and dissecting microscope. Benefit that a compound light microscope provides is that you can observe many kinds of different cells and small organisms while they're still alive. Benefit that a Transmission electron microscope provides is that it is able to produce much greater magnification and shows greater details because of better resolution which optical microscopes can't. Benefit that a Scanning electron microscope provides is that it produces 3-dimensional images.( Does it also have better resolution?) Benefit for a dissecting microscope is?

4. One hand under base while grasping arm with other hand.

5. Put plastic bag over microscope to extend its life.

6. On lab table with arm facing towards you, and its base should be resting evenly on table, approx. 10cm from table's edge.

7.?

8. Coarse adjustment knob

9. Low power?

10. 1500x.

11. 4x = 40x, 10x = 100x, 40x = 400x, 100x = 1000x?

12. Diameter of the circle of light that you see when looking into a microscope. As power gets greater, field of view gets smaller. Lower power = 4.2mm. Medium power = 1.7mm. High power = 0.5mm. (Did an experiment on class about this, so I'm hoping the diameters I got are right.

13. ?

14. High power?

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Sonia

@Miggi

Inactive

Posted: 13 years ago

#524

@ akhl thank u for ur help in the previous questions but could u please solve this one as well?

Engine oil that is classified as SAE 5W-30 has a density of 859kg/m^3 and a viscosity of 6.6Pa.s at a temperture of -30 degree celcius. The oil is poured through a funnel with a neck that is 10.0cm long and 1.00cm diameter. If the level of the oil is kept at a height of 10.0cm above the top of the top of the neck, how long will it take to pour 1.00Liter of oil through the funnel. Hint: YOu may assume that viscous effects need only be considered along the length of the neck of the funnel. Note. Your anaswer will be surprisingly large?

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-InnocentSoul-

@-InnocentSoul-

Sparkler

Posted: 13 years ago

#525

I need help in math plzz help

i gotta factor out polynomials:D

1). x^2-2x-8

2). 10x^2+5x-5

3). 2x^3-7x^2+6x

4). 250x^4-54x

5 40x^4-10x^2

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596277

Suspended

Posted: 13 years ago

#526

Akhl if you have time, could you help me with my biology questions on pg. 71?

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Sonia

@Miggi

Inactive

Posted: 13 years ago

#527

two strings on a violin are stretched between 20cm apart. The tensions of the strings are same. If the ratiocof the frequencies of the fundamental modes is 1.5 to 1.0 then what is the ratio of mass per lengths in the string?

Q: a grand father clock is to be built from a large pendulum so that it will take 1s for the pendullum to go from right to left and 1 more second to go from left to right

A: if the pendulum were to be made of a single point masshow long would it have to be to have the desired period of oscillation?

B: If the bob of the pendulum has 500g mass and it is to oscillate within a 5cm amplitude( 10cm from left to right), what will the energy be? Hint Assume K.E to be zero so only energy at the max amplitude is potential energy.

C: because of air friction and friction on the bearings, the pendulum is found to lose 2% of its enegy for every oscillation cycle. A 5kg mass is attached to a string wound around a pulley of negiligible mass, itself attached to the top pivot of the pendulum, so that the falling mass can provide the energy lost through friction to its steady descent (Assume no acceleration of the falling mass). If total length of the string is 1.1m how many days will it take before pendulum stops oscillating?

Q: A string with mass per unit length mass and length L is under tension T. The string is cut in half and the tension is also halved . The new fundamental frequency of oscillation is?

Edited by Miggi - 13 years ago

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akhl

@akhl

Dazzler

Posted: 13 years ago

#528

@Ayeshaa-
2d) Better to write Iris Diaphragm
2i) Coarse adjustment knob

3) Yes, scanning electron microscope has very high resolution
Dissecting microscope - Easy to view great depth of field

7) At an object on the stage of the microscope. When the objective lenses are changed, then the size of the image of the object changes. Thus, it is possible to know which lenses make image of suitable size.

9) Yes, low power is correct
Fine adjustment is used to focus the image when using the microscope on high power.

13) Depth of field is the thickness of the plane of focus

14) Low power

Note - Better not to use the word "you" in answers.

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akhl

@akhl

Dazzler

Posted: 13 years ago

#529

@Miggi,
two strings on a violin are stretched between 20cm apart. The tensions of the strings are same. If the ratiocof the frequencies of the fundamental modes is 1.5 to 1.0 then what is the ratio of mass per lengths in the string?

Let u = mass per unit length
f1 : f2 = 1.5 : 1.0
Lengths of the string are equal. Therefore the wavelengths for fundamental modes are also equal.
Therefore, v1 : v2 = 1.5 : 1.0
sqrt(T/u1) : sqrt(T/u2) = 1.5 : 1.0
sqrt(u2/u1) = 1.5 / 1.0
u2/u1 = (1.5/1.0)^2 = 2.25/1.00
u1 : u2 = 1.00 : 2.25 (Answer)

Q: a grand father clock is to be built from a large pendulum so that it will take 1s for the pendullum to go from right to left and 1 more second to go from left to right

A: if the pendulum were to be made of a single point masshow long would it have to be to have the desired period of oscillation?

t = 2 pi sqrt(L/g)
t = 2 seconds
Therefore, 2 = 2 pi sqrt(L/g)
sqrt(L/g) = 1/pi
L/g = 1/pi^2
L = g/pi^2 = 1 meter (Answer)

Q: A string with mass per unit length mass and length L is under tension T. The string is cut in half and the tension is also halved . The new fundamental frequency of oscillation is?

f = v/lambda
f = sqrt(T/u) / lambda
T is changed to T/2
lambda is proportional to L. Since L is changed to L/2, therefore lambda is changed to lamda/2
u remains same

Therefore f is changed to
sqrt(1/2) / (1/2) = 2/sqrt(2) = sqrt(2) (Answer)

Will solve other two after some time.