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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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A particle (charge = +0.8 mC) moving in a region where only electric forces act on it has a kinetic energy of 6.7 J at point A.
Maksim231197 [3]

Answer:

The kinetic energy of the particle as it moves through point B is 7.9 J.

Explanation:

The kinetic energy of the particle is:

\Delta K = \Delta E_{p} = q\Delta V

<u>Where</u>:

K: is the kinetic energy

E_{p}: is the potential energy

q: is the particle's charge = 0.8 mC

ΔV: is the electric potential = 1.5 kV                                    

\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J

Now, the kinetic energy of the particle as it moves through point B is:

\Delta K = K_{f} - K_{i}

K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J

Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.

I hope it helps you!      

8 0
4 years ago
A 200kg ball on the end of string is swung in horizontal circle with radius of 0.5m . The ball makes revolution every 2second th
ANEK [815]

Answer:\dfrac{\pi}{2} ms^{-1}

Explanation:

Let T be the time required to make one revolution.

Let r be the radius of the circular path.

Let d be the distance travelled by ball in one revolution.

As we know,the distance travelled in one revolution is the circumference of the circle.

So,d=2\pi r

Given,d=0.5m\\T=2sec

d=2\times \pi \times 0.5=\pi m

Speed of an object moving is circular path is define as the ratio of distance travelled in one revolution to the time taken by the object to complete one revolution.

Let s be the speed of the ball.

s=\frac{d}{T}=\frac{\pi }{2}ms^{-1}

So,the speed of the ball is \frac{\pi }{2}ms^{-1}

5 0
3 years ago
Types of media that can influence one's social development include all of the following except
dybincka [34]
A. movies b. internet c. books
6 0
3 years ago
Read 2 more answers
0.042 kg … ________________grams
olchik [2.2K]

Answer:

42 grams

Explanation:

you multiply the value by 1000 to get your answer

6 0
3 years ago
Read 2 more answers
7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
wlad13 [49]

Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

3 0
3 years ago
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