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3 years ago

10

A camera lens focuses on an object 75.0 cm from the lensThe image forms 3.50 cm behind the lens. What is the magnification of th

e image?

1 answer:

N76 [4]3 years ago

6 0

Answer:

7/150

Explanation:

The following data were obtained from the question:

Object distance (u) = 75cm

Image distance (v) = 3.5cm

Magnification (M) =..?

Magnification is simply defined as:

Magnification (M) = Image distance (v)/ object distance (u)

M = v /u

With the above formula, we can obtain the magnification of the image as follow:

M = v/u

M = 3.5/75

M = 7/150

Therefore, the magnification of the image is 7/150.

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Imagine a 15 kg block moving with a velocity of 20 m/s to the left. Calculate the kinetic Energy of this block.

ivann1987 [24]

Answer:

3000 J

Explanation:

Kinetic energy is:

KE = ½ mv²

If m = 15 kg and v = -20 m/s:

KE = ½ (15 kg) (-20 m/s)²

KE = 3000 J

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2 years ago

A vase falls from a windowsill 50.0 m above the sidewalk. How fast is the base moving when it hits the ground?

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Answer:

The answer is A sorry if i'm wrong

Explanation:

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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

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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>

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

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

$\Delta h=0.0835\ m$

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2 years ago

A weight of 1400 pounds is suspended from two cables as shown in the figure. What is the tension in the left cable? _________ po

juin [17]

Answer:

Following are the solution to this question:

Explanation:

Law:

$\to \theta= 180^{\circ}- 50^{\circ}- 25^{\circ}$

$= 180^{\circ}- 75^{\circ}\\\\= 105^{\circ}$

$\to \frac{T_{L}}{\sin (90+50)}= \frac{T_{R}}{\sin (25+90)}=\frac{1400}{\sin (105)}$

$\to T_L=931.65 \ pounds \\\\ \to T_R=1313.59 \ pounds \\\\$

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2 years ago

Suppose light from a 632.8 nm helium-neon laser shines through a diffraction grating ruled at 520 lines/mm. How many bright line

Leya [2.2K]

Answer:

1 bright fringe every 33 cm.

Explanation:

The formula to calculate the position of the m-th order brigh line (constructive interference) produced by diffraction of light through a diffraction grating is:

$y=\frac{m\lambda D}{d}$

where

m is the order of the maximum

$\lambda$ is the wavelength of the light

D is the distance of the screen

d is the separation between two adjacent slit

Here we have:

$\lambda=632.8 nm = 632.8\cdot 10^{-9} m$ is the wavelength of the light

D = 1 m is the distance of the screen (not given in the problem, so we assume it to be 1 meter)

$n=520 lines/mm$ is the number of lines per mm, so the spacing between two lines is

$d=\frac{1}{n}=\frac{1}{520}=1.92\cdot 10^{-3} mm = 1.92\cdot 10^{-6} m$

Therefore, substituting m = 1, we find:

$y=\frac{(632.8\cdot 10^{-9})(1)}{1.92\cdot 10^{-6}}=0.330 m$

So, on the distant screen, there is 1 bright fringe every 33 cm.

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3 years ago