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4 years ago

Calculate the limit of resolution for the oil lens of your microscope. assume an average wavelength of 500 nm

1 answer:

3 0

The resolution of a microscope is the distance with the shortest measurement between two different points given a specimen with the premise that it can still be seen clearly or distinguished by the one looking through the microscope. It can be calculate from the ratio of the wavelength of the light and twice the numerical aperture or the refractive index of the lens. Most of the microscopes have a numerical aperture ranging from 1.2 to 1.4. Resolution and the numerical aperture are indirectly proportional so that as the aperture increases the resolution would decrease. We calculate as follows:

<span>Resolution = wavelength / ((2) (numerical aperture))
Resolution = 500 nm / (2 ) ( 1.25) = 200 nm = 0.2 um</span>

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7. These temperatures were recorded in Pasadena for a week in April. 87 85 80 78 83 86 90 Find each of these. (a) Mean (e) Range

slega [8]

Answer:

a) Mean = 84.14

b) Median = 85

c) Mode = no mode (since there is no variable that appears more than once in this dataset)

d) Midrange = 84

e) Range = 12

f) Variance = 14.69

g) Standard deviation = 3.83

Explanation:

The raw data to be processed is

87 85 80 78 83 86 90

a) Mean = (Σx)/N

The mean is the sum of variables divided by the number of variables

x = each variable

N = number of variables = 7

Mean = (87+85+80+78+83+86+90)/7

Mean = 84.14

b) Median is the number in the middle of the dataset when the variables are arranged in ascending or descending order.

Arranging the data in ascending order

78, 80, 83, 85, 86, 87, 90

The number in the middle is the 4th number = 85

Median = 85

c) Mode is the variable that occurs the most in a distribution.

For this question, all of the variables occur only once, with no variable occurring more than once. Hence, there is no mode for this dataset.

d) Midrange is the arithmetic mean of the highest and lowest number in the dataset.

Mathematically,

Midrange = (Highest + Lowest)/2

Midrange = (90 + 78)/2

Midrange = 84

e) Range is the difference the highest and the lowest numbers in a dataset.

Range = 90 - 78 = 12

f) Variance is an average of the squared deviations from the mean.

Mathematically,

Variance = [Σ(x - xbar)²/N]

xbar = mean

Σ(x - xbar)² = (78 - 84.14)² + (80 - 84.14)² + (83 - 84.14)² + (85 - 84.14)² + (86 - 84.14)² + (87 - 84.14)² + (90 - 84.14)² = 102.8572

Variance = (102.8572)/7

Variance = 14.69

g) Standard deviation = √(variance)

Standard deviation = √(14.69)

Standard deviation = 3.83

Hope this Helps!!!

8 0

4 years ago

Read 2 more answers

What creates more pressure?Check all that apply A.Gas molecules moving more quickly B.More gas molecules C. Going to a higher al

Shkiper50 [21]

It's A.Gas molecules moving more quickly..... hope it helps you dear

7 0

3 years ago

What is a stream’s discharge rate if it has a width of 10 meters, a depth of 2 meters, and a velocity of 2 meters per second?

julia-pushkina [17]

A stream’s discharge rate if it has a width of 10 meters, a depth of 2 meters, and a velocity of 2 meters per second will be 40 $m^{3}$ /sec .

Discharge rate = velocity * area

= velocity * depth * width

= 2 * 2 * 10 = 40 $m^{3}$ /sec

A stream’s discharge rate if it has a width of 10 meters, a depth of 2 meters, and a velocity of 2 meters per second will be 40 $m^{3}$ /sec .

learn more about discharge rate

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4 0

2 years ago

A current-carrying wire oriented north-south and laid over a compass deflects the compass 8° to the east. What is the magnitude

zhenek [66]

$2.8 \times 10^{-6}\ T$ is the magnitude of the magnetic field made by the current

<u>Explanation:</u>

Given data:

$\theta=8^{\circ}$

Magnetic field of earth, $B_{\text {earth}}=2 \times 10^{-5}\ \text {tesla}$

We need to find the magnetic field of wire, $B_{\text {wire }}$

The compass needle moves toward a direction of magnetic field. The current in wire makes a magnetic field in available space where the compass is on the ground. The vector sum of the Earth's magnetic field and the wire's magnetic field represents the net magnetic field, as shown in the attached drawing, expressing the angle:

$\tan \theta=\frac{B_{\text {wire}}}{B_{\text {earth}}}$

By substituting the given values, we get

$B_{\text {wire}}=\tan \theta \times B_{\text {earth}}=\tan 8^{\circ} \times 2 \times 10^{-5}=0.1405 \times 2 \times 10^{-5}$

$B_{\text {wire}}=0.28 \times 10^{-5}=2.8 \times 10^{-6}\ T$

4 0

3 years ago

If launched at the right speed, what could eventually happen to objects launched from Newton's theoretical cannon on the top of

Nadusha1986 [10]

If the speed is higher than the orbital velocity, but not high enough to leave Earth altogether (lower than the escape velocity), it will continue revolving around Earth along an elliptical orbit. (D) for example horizontal speed of 7,300 to approximately 10,000 m/s for Earth.

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3 years ago