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3 years ago

How does the thickness of the lens affects its focal length plz ans this i will mark as brainliest

Physics

1 answer:

Oksanka [162]3 years ago

5 0

Hello! :)

The focal length of the lens tells you how far away from the lens a focused image is created, if light rays approaching the lens are parallel. A lens with more “bending power” has a shorter focal length, because it alters the path of the light rays more effectively than a weaker lens. Most of the time, you can treat a lens as being thin and ignore any effects from the thickness, because the thickness of the lens is much less than the focal length. But for thicker lenses, how thick they are does make a difference, and in general, results in a shorter focal length.

Hope I helped and didn’t answer too late!

Good luck and stay COOL!

~ Destiny ^_^

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A 250 kg flatcar 25 m long is moving with a speed of 3.0 m/s along horizontal frictionless rails. A 61 kg worker starts walking

Anna11 [10]

Answer:

x=31.09m

Explanation:

p1=p2

The momentum of flatcar and the momentum of the worker so

The velocity of the worker is:

$m_{f}*v_{f}=m_{w}*v_{w}\\\\v_{f}=\frac{m_{f}*v_{f}}{m_{w}}\\v_{f}=\frac{61kg*3.0\frac{m}{s}}{250kg}\\v_{f}=0.732\frac{m}{s}$

The total motion has a total velocity and is

$Vt=v_{w}+v_{f}\\Vt=0.732\frac{m}{s}+3.0\frac{m}{s}\\Vt=3.732\frac{m}{s}$

The time the worker take walking is

$t=\frac{x}{v_{w}}\\t=\frac{25m}{3\frac{m}{s}}=8.33s$

Now the total time and the total velocity determinate the motion of tha flatcar how far has moved

$x=t*Vt\\x=8.33s*3.732\frac{m}{s} \\x=31.09m$

5 0

2 years ago

You connect a 100-resistor, a 800-mH inductor, and a 10.0-uF capacitor in series across a 60.0-Hz, 120-V (peak) source. The impe

steposvetlana [31]

Answer:

Impedance, Z = 107 ohms

Explanation:

It is given that,

Resistance, R = 100 ohms

Inductance, $L=800\ mH=800\times 10^{-3}\ H=0.8\ H$

Capacitance, $C=10\ \mu F=10\times 10^{-6}\ F=10^{-5}\ F$

Frequency, f = 60 Hz

Voltage, V = 120 V

The impedance of the circuit is given by :

$Z=\sqrt{R^2+(X_C-X_L)^2}$ ...........(1)

Where

$X_C$ is the capacitive reactance, $X_C=\dfrac{1}{2\pi fC}$

$X_C=\dfrac{1}{2\pi \times 60\times 10^{-5}}=265.65\ \Omega$

$X_L$ is the inductive reactance, $X_L={2\pi fL}$

$X_L={2\pi \times 60\times 0.8}=301.59\ \Omega$

So, equation (1) becomes :

$Z=\sqrt{(100)^2+(265.65-301.59)^2}$

Z = 106.26 ohms

Z = 107 ohms

So, the impedance of the circuit is 107 ohms. Hence, this is the required solution.

8 0

3 years ago

A system gains 767 kJ of heat, resulting in a change in internal energy of the system equal to +151 kJ. How much work is done?

Crazy boy [7]

Answer:

The work done on the system is -616 kJ

Explanation:

Given;

Quantity of heat absorbed by the system, Q = 767 kJ

change in the internal energy of the system, ΔU = +151 kJ

Apply the first law of thermodynamics;

ΔU = W + Q

Where;

ΔU is the change in internal energy

W is the work done

Q is the heat gained

W = ΔU - Q

W = 151 - 767

W = -616 kJ (The negative sign indicates that the work is done on the system)

Therefore, the work done on the system is -616 kJ

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2 years ago

A stone dropped from the top of a 80m high building strikes the ground at 40 m/s after falling for 4 seconds. The stone's potent

Sergeu [11.5K]

Answer: c. 40m

Explanation:

See picture

3 0

3 years ago

The table and graph below show the distances traveled by two different objects. (3 points)

Aleks04 [339]

Answer:c

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5 0

2 years ago