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Tomtit [17]
2 years ago
9

A camera lens with focal length f = 50 mm and maximum aperture f>2

Physics
1 answer:
Brut [27]2 years ago
8 0

Answer:

The minimum distance between two points on the  object that are barely resolved is 0.26 mm

The corresponding distance between the  image points = 0.0015 m

Explanation:

Given  

focal length f = 50 mm and maximum aperture f>2

s =  9.0 m

aperture = 25 mm = 25 *10^-3 m

Sin a = 1.22 *wavelength /D  

Substituting the given values, we get –  

Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m

Sin a = 2.93 * 10 ^-5 rad

Now  

Y/9.0 m = 2.93 * 10 ^-5

Y = 2.64 *10^-4 m = 0.26 mm

Y’/50 *10^-3 = 2.93 * 10 ^-5  

Y’ = 0.0015 m

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A rope of negligible mass passes over a uniform cylindrical pulley of 1.50kg massand 0.090m radius. The bearings of the pulley h
boyakko [2]

Answer:

Explanation:

mass of pulley, m3 = 1.5 kg

Radius of pulley, R = 0.09 m

mass of monkey, m2 = 4.5 kg

mass of banana bunch, m1 = 3 kg

Let a is teh acceleration ans T1 and T2 be the tension in the rope.

The moment of inertia of the pulley

I = 0.5 x m3 x R² = 0.5 x 1.5 x 0.09 x 0.09 = 0.006075 kgm²

According to Newton's second law

T1 - m1 g = m1 x a .... (1)

m2 g - T2 = m2 x a ..... (2)

(T2 - T1 ) x R = I x α    

where, α is the angular acceleration

α = a / R

(T2 - T1)R = 0.5 x m3 x R² x a / R

T2 - T1 = 0.5 x m3 x a ..... (3)  

from (1), (2) and (3)

a = \frac{m_{2}-m_{1}}{m_{1}+m_{2}+\frac{m_{3}}{2}}\times g

a = \frac{4.5-3}{3+4.5+0.75}\times 9.8

a = 1.78 m/s²

from equation (1)

T1 = m1 ( g + a) = 3 ( 9.8 + 1.78) = 34.77 N

from equation (2)

T2 = m2 (g - a) = 4.5 (9.8 - 1.78) = 36.13 N

4 0
3 years ago
A +0.2 µC charge is in an electric field. What happens if that charge is replaced by a +0.4 µC charge?
AleksandrR [38]

Explanation:

It is known that electric field is responsible for creating electric potential. As a result, it depends only on the electric field and not on the magnitude of charge.

So, when a charge is increased by a factor of 2 then electric potential will remain the same. Since, expression to calculate the electric potential is as follows.

                 U = qV

Since, the electric potential is directly proportional to the charge. Hence, when 0.2 \mu C tends to replaced by 0.4 \mu C then charge is increased by a factor of 2. Hence, the electric potential energy is doubled.

Thus, we can conclude that if that charge is replaced by a +0.4 µC charge then electric potential stays the same, but the electric potential energy doubles.

4 0
3 years ago
An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get
Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

mV_i = (m-m_O)V_f - m_OV_O

where,

m=Total mass

m_O = Mass of Object

V_i = Velocity before throwing

V_f = Final Velocity

V_O = Velocity of Object

Our values are:

m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg

Solving to find the final speed, after throwing the object we have

V_f=\frac{mV_0+m_TV_O}{m-m_O}

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) 5.3Kg\rightarrow 15m/s

V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}

V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}

V_{f1}=0.8511m/s

B) 7.9Kg\rightarrow 11.2m/s

V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}

V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}

V_{f2} = 2.0173m/s

C) 10.5Kg\rightarrow 7m/s

V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}

V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}

V_{f3} = 3.63478m/s

Therefore the final velocity of astronaut is 3.63m/s

7 0
3 years ago
Is the force of friction the sum of all the microscopic contact forces between the bottom of the block and the surface?
kotegsom [21]

The force of friction is that force that tends to oppose the motion of the body with the surface in contact... are you clear with it?

6 0
2 years ago
Un protón se acelera por una diferencia de potencial de 2.6k V y se dirige a una región donde existe un campo magnético uniforme
Bond [772]

Answer:

5.08*10^{-13}N

Explanation:

With this information we can calculate the velocity of the proton by taking into account the kinetic energy of the proton:

qV=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(1.6*10^{-19}C)(2.6*10^3V)}{1.67*10^{-27}kg}}=705835.3m/s

The magnitude of the magnetic force will be:

F=qvB=(1.6*10^{-19}C)(705835.3m/s)(4.5T)=5.08*10^{-13}N

hope this helps!

7 0
3 years ago
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