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3 years ago

A 3.8 kg object is lifted 12 meters. approximately how much work is performed during the lifting.

Physics

2 answers:

Dmitry [639]3 years ago

7 0

work done=446.9 J . so option (c) is correct.

Explanation:

the formula for work done is given by

W= F d

F= force= mg where m= mass and g= acceleration due to gravity

F= 3.8 (9.8)=37.24 J

so W=37.24 (12)

W=446.9 J

zimovet [89]3 years ago

6 0

Answer:

Explanation:

446.9 j

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The distance between the first and fifth minima of a single-slit diffraction pattern is 0.350 mm with the screen 39.0 cm away fr

rjkz [21]

Answer: a) the width is 2.54 mm

b) the angle is 0.012°

Explanation:

we have the equation:

λ = zw/(L*m)

Where is the distance for the middle of the screen, L is the distance to the screen, w is the widht of the slit and λ is the wavelength.

now we can isolate z and get:

z = λL*m/w

and the distance between the 5 minimum and the first one is 0.035mm

then:

0.035mm = λL*5/w - λL*1/w = λL*4/w

now we can solve it for w.

0.35mm = 570nm*39.0cm*4/w

now, we have all diferent units, lets use nanometers.

1cm = 1x10^-2 m

1mm = 1x10^-3 m

1nm = 1x10^-9m

then: 1 cm = 1x10^7 nm

1 mm = 1x10^6 nm

then we have:

0.35x10^6 nm = 570nm*39.0x10^7 nm*4/w

w = (570nm*39.0x10^7 nm*4)/0.35x10^6 nm =(570nm*39.0*10*4)/0.35

w = 2540571.4 nm

it is a bigg number, let's write it in milimeters:

w = (2540571.4/10^6) mm = 2.54 mm

the first minimum can be obtained by the equation w*sinθ = mλ by using m = 1

then we have:

2540571.4 nm*sinθ = 570 nm

sinθ = 570/2540571.4 = 0.000224

θ = asin(0.00022) = 0.012°

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Answer:
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4 years ago

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An object with a height of 4.0 cm is placed 30.0 cm from a lens. The resulting inverted image has a height of 1.5 cm. What is th

jarptica [38.1K]

Answer:

Focal length of the lens is 8.2 cm.

Explanation:

It is given that,

Height of object, h = 4 cm

Object distance, u = -30 cm

Height of the image, h' = -1.5 cm (negative because the image is inverted)