If you could help me out that would be fantastic.?!!!

A. Calculate the diffraction limit of the human eye, assuming a wide-open pupil so that your eye acts like a lens with diameter

myrzilka [38]

A. The Dawes limit tells us that the resolving power is equal to 11.6 / d, where d is the diameter of the eye’s pupil in units of centimeters. The eye's pupil can dialate to approximately 7 mm, or 0.7 cm. So 11.6 / .7 = 16.5 arc seconds, or about a quarter arc minute ~ 17 arc seconds

Although, the standard answer for what people can really see is about 1 arc minute.

B. It is considered as linear, so given a 10 meter telescope (10,000 mm):

10000 / 7 = 1428 times better for the 10 meter scope ~ 1400 times better (in 2 significant figures)

C. For a 7 cm interferometer, that is just similar to a 7 cm scope. Therefore we would expect

11.6 / 7 = 1.65 arc seconds ~ 1.7 arc seconds

This value is what we typically can get from a 7 cm scope.

6 0

2 years ago

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If a questions says "at what point does the ball stop?" what is it asking for?

Bad White [126]

Explanation:

If a question says "at what point does the ball stop?", it means we need to find the position of the ball when its final velocity is equal to 0. It can be calculated using the equation of kinematics as follows :

d = ut + (1/2) at²

and

v²-u²=2ad

Where, u is initial velocity, v is final velocity, a is acceleration, t is time and d is displacement.

5 0

2 years ago

What does “use f=ma, where a=v-u divided by t”mean

m_a_m_a [10]

$f = m( \frac{v - u}{t} )$

4 0

2 years ago

Two charged concentric spherical shells have radii of 11.0 cm and 14.0 cm. The charge on the inner shell is 3.50 ✕ 10−8 C and th

Sergio039 [100]

Answer:

The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

Explanation:

Given that,

Radius of inner shell = 11.0 cm

Radius of outer shell = 14.0 cm

Charge on inner shell $q_{inn}=3.50\times10^{-8}\ C$

Charge on outer shell $q_{out}=1.60\times10^{-8}\ C$

Suppose, at r = 11.5 cm and at r = 20.5 cm

We need to calculate the magnitude of the electric field at r = 11.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Where, q = charge

k = constant

r = distance

Put the value into the formula

$E=\dfrac{9\times10^{9}\times3.50\times10^{-8}}{(11.5\times10^{-2})^2}$

$E=2.38\times10^{4}\ N/C$

The total charge enclosed by a radial distance 20.5 cm

The total charge is

$q=q_{inn}+q_{out}$

Put the value into the formula

$q=3.50\times10^{-8}+1.60\times10^{-8}$

$q=5.1\times10^{-8}\ C$

We need to calculate the magnitude of the electric field at r = 20.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times5.1\times10^{-8}}{(20.5\times10^{-2})^2}$

$E=1.09\times10^{4}\ N/C$

Hence, The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

7 0

2 years ago

The Force of friction between an object and the surface upon which it is sliding is 12N. The weight of the object is 20N. What i

VashaNatasha [74]

Answer:

Explanation:

Reducing Sliding Friction. You can reduce the resistive force of sliding friction by applying lubrication between the two surfaces in contact, by using rollers, or by decreasing the normal force

6 0

2 years ago