PLEASE ASSIST 20 POINTS

When we move from 100x total magnification to 400x total magnification on a compound light microscope, what happens with resolut

gayaneshka [121]

1. a) In order to sharpen the objective with a 40x objective (400x total magnification), the objective must be closer to the sample than with a 10x objective (100x total magnification), so the sample is out of focus.

b) At 400x total magnification, the field of view is narrower than at 100x total magnification.

c) A 40x objective (i.e. 400x total magnification) has a higher numerical aperture than a 10x objective (i.e. 100x total magnification) and therefore provides better resolution.

d) Shallow depth of field.

e) The amount of light is reduced.

f) Shorter working distance.

2. The best objective for observing bacteria is the 100x objective, which is the immersion objective.

3. Without the addition of immersion oil, the sample will look blurry and will not give maximum resolution.

Learn more about sharpen here;

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5 0

2 years ago

Calculate specific heat of liquid if a 450. g iron block is heated to 98.0 C and then droped into a calorimeter with 125 g of un

Naya [18.7K]

Answer:

Specific heat of liquid $1.258 \ J/g^0C.$

Explanation:

We know in thermal equilibrium :

Loss in heat by iron block = Gain in heat by liquid .

Specific heat of iron = 0.45 $J/g^0C.$ { source internet }

Now , loss in heat by iron block = $mC_{iron}\Delta T=450\times 0.45\times 57.6=11664\ J.$

Heat gain by liquid= $mC_{liquid}\Delta T=450\times C_{liquid}\times 20.6=9270\times C_{liquid}.$

Equating both we get :

$C_{liquid}=1.258 \ J/g^0C.$

7 0

3 years ago

3. If a pipe with flowing water has a cross-sectional area nine times greater at point 2 than at point 1, what would be the rela

shusha [124]

Answer:

The velocity of the flow at point 1 is nine times greater than the velocity at point 2.

Explanation:

We know that the same volume of water entering through point 1 must exit through point 2.

The volume of water per unit of time or the volumetric flow is defined by:

$V_{flow}=v*A\\ where:\\V_{flow}= flow [m^3/s]\\v=velocity[m/s]\\A=area [m^2]$

If $A_{2}=9*A_{1} \\$

therefore

$v_{1} *A_{1} = v_{2} *A_{2} \\v_{1} *A_{1} = v_{2} *9A_{1} \\\\v_{1} = 9v_{2} \\$

4 0

4 years ago

At one instant a bicyclist is 21.0 m due east of a park's flagpole, going due south with a speed of 13.0 m/s. Then 21.0 s later,

andre [41]

Answer:

Part a)

$d = 21\sqrt2 = 29.7 m$

Part b)

Direction is 45 degree North of West

Part c)

$v_{avg} = 1.41 m/s$

Part d)

direction of velocity will be 45 Degree North of West

Part e)

$a = 0.875 m/s^2$

Part f)

$\theta = 45 degree$ North of East

Explanation:

Initial position of the cyclist is given as

$r_1 = 21.0 m$ due East

final position of the cyclist after t = 21.0 s

$r_2 = 21.0 m$ due North

Part a)

for displacement we can find the change in the position of the cyclist

so we have

$d = r_2 - r_1$

$d = 21\hat j - 21\hat i$

so magnitude of the displacement is given as

$d = 21\sqrt2 = 29.7 m$

Part b)

direction of the displacement is given as

$\theta = tan^{-1}\frac{y}{x}$

$\theta = tan^{-1}\frac{21}{-21}$

so it is 45 degree North of West

Part c)

For average velocity we know that it is defined as the ratio of displacement and time

so here the magnitude of average velocity is defined as

$v_{avg} = \frac{\Delta x}{t}$

$v_{avg} = \frac{29.7}{21}$

$v_{avg} = 1.41 m/s$

Part d)

As we know that average velocity direction is always same as that of average displacement direction

so here direction of displacement will be 45 Degree North of West

Part e)

Here we also know that initial velocity of the cyclist is 13 m/s due South while after t = 21 s its velocity is 13 m/s due East

So we have

change in velocity of the cyclist is given as

$\Delta v = v_f - v_i$

$\Delta v = 13\hat i - (-13\hat j)$

now average acceleration is given as

$a = \frac{\Delta v}{\Delta t}$

$a = \frac{13\hat i + 13\hat j}{21}$

so the magnitude of acceleration is given as

$a = \frac{13\sqrt2}{21} = 0.875 m/s^2$

Part f)

direction of acceleration is given as

$\theta = tan^{-1}\frac{y}{x}$

$\theta = tan^{-1}\frac{13}{13}$

$\theta = 45 degree$ North of East

4 0

3 years ago

When a viscous fluid flows in a tube, its velocity is __________.

ankoles [38]

From the equation, r is the distance from the center of the tube. When r = 0, you're considered to be at the centerline and the expression has the highest value. Therefore, the velocity of a viscous fluid is greatest at the center of the tube.

4 0

3 years ago