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Law Incorporation [45]

2 years ago

12

You place an object 20 cm from a lens and find an image on the opposite side 30 cm from the lens. You calculate a focal length o

f 12 cm. You then move the object 30 cm from the same lens and once again measure the image distance. All your measurements and calculations are correct. What should you find when you calculate this focal length? a) It increased. b) It decreased. c) It stayed the same. d) There is not enough information to determine this focal length.

1 answer:

Bezzdna [24]2 years ago

7 0

Answer:

The correct option is 'c': It stays same.

Explanation:

The focal length of any lens is obtained by using lens maker's formula as follows

$\frac{1}{f}=(\frac{n_{2}}{n_{2}}-1)(\frac{1}{R_{1}}-\frac{1}{R_{2}})$

where

$n_{2}$ is the refractive index of the material of lens

$n_{1}$ is the refractive index of the medium surrounding the lens

$R_{1},R_{2}$ are the radii of the 2 surfaces of the lens.

As we see that in the formula the focal length of the lens only depends on the refractive indexes and the radius of the lens thus we conclude that the focal length of the lens is independent on the position of object or image.

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Which type of light wave has the highest energy? which has the least amount of energy

devlian [24]

Answer:

Red has the lowest energy and violet the highest. Beyond red and violet are many other kinds of light our human eyes can't see, much like there are sounds our ears can't hear. On one end of the electromagnetic spectrum are radio waves, which have wavelengths billions of times longer than those of visible light.

Explanation:

6 0

2 years ago

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2. Secret agent Austin Powers jumps on ju

denpristay [2]

Answer:

a) $h=250\ m$

b) $\Delta h=0.0835\ m$

Explanation:

Given:

upward acceleration of the helicopter, $a=5\ m.s^{-2}$

time after the takeoff after which the engine is shut off, $t_a=10\ s$

a)

<u>Maximum height reached by the helicopter:</u>

using the equation of motion,

$h=u.t+\frac{1}{2} a.t^2$

where:

u = initial velocity of the helicopter = 0 (took-off from ground)

t = time of observation

$h=0+0.5\times 5\times 10^2$

$h=250\ m$

b)

time after which Austin Powers deploys parachute(time of free fall), $t_f=7\ s$

acceleration after deploying the parachute, $a_p=2\ m.s^{-2}$

<u>height fallen freely by Austin:</u>

$h_f=u.t_f+\frac{1}{2} g.t_f^2$

where:

$u=$ initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

$t_f=$ time of free fall

$h_f=0+0.5\times 9.8\times 7^2$

$h_f=240.1\ m$

<u>Velocity just before opening the parachute:</u>

$v_f=u+g.t_f$

$v_f=0+9.8\times 7$

$v_f=68.6\ m.s^{-1}$

<u>Time taken by the helicopter to fall:</u>

$h=u.t_h+\frac{1}{2} g.t_h^2$

where:

$u=$ initial velocity of the helicopter just before it begins falling freely = 0

$t_h=$ time taken by the helicopter to fall on ground

$h=$ height from where it falls = 250 m

now,

$250=0+0.5\times 9.8\times t_h^2$

$t_h=7.1429\ s$

From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.

<u>remaining time,</u>

$t'=t_h-t_f$

$t'=7.1428-7$

$t'=0.1428\ s$

<u>Now the height fallen in the remaining time using parachute:</u>

$h'=v_f.t'+\frac{1}{2} a_p.t'^2$

$h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2$

$h'=9.8165\ m$

<u>Now the height of Austin above the ground when the helicopter crashed on the ground:</u>

$\Delta h=h-(h_f+h')$

$\Delta h=250-(240.1+9.8165)$

$\Delta h=0.0835\ m$

5 0

2 years ago

A system had 150 kj of work done on it and its internal energy increased by 60 kj. How much energy did the system gain or lose a

mina [271]

Answer:

The system loses 90 kJ of heat

Explanation:

We can answer the question by using the 1st law of thermodynamics, which states that:

$\Delta U=Q-W$

where

$\Delta U$ is the change in internal energy of the system

$Q$ is the heat absorbed by the system (positive if absorbed, negative if released by the system)

$W$ is the work done by the system (positive if done by the system, negative if done by the surrounding on the system)

In this problem, we have:

$W=-150 kJ$ is the work done (negative, because it is done by the surrounding on the system)

$\Delta U=+60 kJ$ is the increase in internal energy

Using the equation above, we can find Q, the heat absorbed/released by the system:

$Q=\Delta U+W=+60 kJ+(-150 kJ)=-90 kJ$

And the negative sign means that the system has lost this heat.

8 0

3 years ago

Hi..Please answer questions???

Nadya [2.5K]

Answer:

A

Explanation:

Becuse its complete number

3 0

3 years ago

A steel cable has a cross-sectional area 4.49 × 10^-3 m^2 and is kept under a tension of 2.96 × 10^4 N. The density of steel is

Lemur [1.5K]

Answer:

The transverse wave will travel with a speed of 25.5 m/s along the cable.

Explanation:

let T = 2.96×10^4 N be the tension in in the steel cable, ρ = 7860 kg/m^3 is the density of the steel and A = 4.49×10^-3 m^2 be the cross-sectional area of the cable.

then, if V is the volume of the cable:

ρ = m/V

m = ρ×V

but V = A×L , where L is the length of the cable.

m = ρ×(A×L)

m/L = ρ×A

then the speed of the wave in the cable is given by:

v = √(T×L/m)

= √(T/A×ρ)

= √[2.96×10^4/(4.49×10^-3×7860)]

= 25.5 m/s

Therefore, the transverse wave will travel with a speed of 25.5 m/s along the cable.

7 0

3 years ago