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Rufina [12.5K]
3 years ago
12

A microscope using ultraviolet light is used to study bacteria. If the aperture diameter is 1.5 cm and it is desired to distingu

ish features with an angular size 0.036 arc seconds, what maximum wavelength can be used
Physics
1 answer:
Hatshy [7]3 years ago
3 0

Given :

A microscope using ultraviolet light is used to study bacteria. If the aperture diameter is 1.5 cm.

Angular size is 0.036 arc seconds.

To Find :

The maximum wavelength that can be used.

Solution :

Converting given angle into radians :

\theta = \dfrac{0.036}{3600}\times \dfrac{\pi}{180}\\\\\theta = 1.745\times 10^{-7}\ radians

Now, we know maximum wavelength is given by :

\lambda = \dfrac{\theta \times D}{1.22}\\\\\lambda = \dfrac{1.745\times 10^{-7} \times 0.015}{1.22}\ m\\\\\lambda =2.145 \times 10^{-7}\ m

Hence, this is the required solution.

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7 0
2 years ago
A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik
Nina [5.8K]

Answer:

1 × 10⁶ N/C

Explanation:

The magnitude of the electric field between the membrane = surface density / permittivity of free space = 10 ⁻⁵C/ m² / (8.85 × 10⁻¹²N⁻¹m⁻²C²) = 1.13 × 10⁶ N/C approx 1 × 10⁶ N/C

4 0
3 years ago
In a football game, the running back takes a handoff and begins running toward midfield at 3.21 yards/s . As he moves through hi
Inessa [10]

Given Information:

Initial speed = u = 3.21 yards/s

Acceleration = α  = 1.71 yards/s²

Final speed = v = 7.54 yards/s

Required Information:

Distance = s = ?

Answer:

Distance = s = 13.61

Explanation:

We are given the speeds and acceleration of the runner and we want to find out how much distance he covered before being tackled.

We know from the equations of motion,

v² = u² + 2αs

Where u is the initial speed of the runner, v is the final speed of the runner, α is the acceleration of the runner and s is the distance traveled by the runner.

Re-arranging the above equation for distance yields,

2αs = v² - u²

s = (v² - u²)/2α

s = (7.54² - 3.21²)/2×1.71

s = 46.55/3.42

s = 13.61 yards

Therefore, the runner traveled a distance of 13.61 yards before being tackled.

8 0
3 years ago
. Water is flowing at 12m/s in a horizontal pipe under a pressure of 600kpa radius 2cm. a. What is the speed of the water on the
lana66690 [7]

Answer:

Outlet Velocity = 192 m/s

Outlet Pressure = 510 kPa

Explanation:

Givens:

Inlet Velocity, V₁ = 12 m/s

Inlet Pressure, P₁ = 600 kPa = 600,000 Pa

Inlet Radius r₁ = 0.5 cm

Outlet Velocity, V₂ = not given (we are asked to find this)

Outlet Pressure,  P₂ = not given (we are asked to find this)

Outlet Radius, r₂ = 0.5 cm

From these, we can find the following:

Inlet Area, A₁ = π (r₁)² = π(2)² = 4π cm²

Outlet Area, A₂ = π (r₂)² = π(0.5)² = 0.25π cm²

<u>Part A :</u>

Assuming that water is incompressible, we can reason that within the same given time, the amount of volume of water entering the inlet must equal the volume of water exiting the outlet. Hence by the continuity equation (i.e. conservation of mass)

Inlet Volume flow rate = Outlet Volume flow rate

(recall that Volume flow rate in a pipe is given by Velocity x Cross Section Area), Hence the equation becomes

V₁ x A₁ = V₂ x A₂  (substituting the values that we know from above)

12 x 4π = V₂ x 0.25π  (we don't have to change all to SI units because the conversion factors on the left will cancel out the conversion factors on the right).

V₂ = (12 x 4π) / (0.25π)

V₂ = 192 m/s  (Answer)

<u>Part B:</u>

For Part B, if we assume a closed ideal system (control volume method), we can simply apply the energy equation (i.e Bernoulli's equation)

P₁ + (1/2)ρV₁ + ρgh₁ = P₂ + (1/2)ρV₂ + ρgh₂

Because the pipe is horizontal, there is no difference between h₁ and h₂, hence we can neglet this term:

P₁ + (1/2)ρV₁ = P₂ + (1/2)ρV₂  (rearranging)

P₂ = P₁ + (1/2)ρV₁ - (1/2)ρV₂

= P₁  + (1/2)ρ (V₁-V₂)

Assuming that the density of water is approx, ρ = 1000 kg/m³

P₂ = 600,000  + (1/2)(1000) (12-192)

= 600,000  + ( -90,000)

= 510,000 Pa

= 510 kPa (Answer)

8 0
3 years ago
Lightning can sometimes occur on hot and humid summer evenings when there are no thunderstorms.
mart [117]

Answer:

(B) False

Explanation:

No, it is not possible to have thunder without lightning. Thunder is a direct result of lightning.

4 0
3 years ago
Read 2 more answers
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