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amm1812
3 years ago
10

Read each example and identify whether the data are observations or inferences from observations. The fish’s ventral fin measure

d 8.5 cm long.
Physics
1 answer:
drek231 [11]3 years ago
4 0
8.5 cm long just bc we love this
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Please answer this question given in the picture
elena-s [515]

A plane mirror forms a virtual image behind the mirror. The image is as far behind the mirror as the object is in front of it. A cannot see his image because the length of the mirror is too short on his side. However, he can see the objects placed at points P and Q, but cannot see the object placed at point R



Hope this helps Buddy!


~ Courtney

6 0
4 years ago
On April 15, 1999, a South Korean cargo plane crashed due to a confusion over units. The plane was to fly from Shanghai, China,
sattari [20]

Answer:

993 m or 3257 ft

Explanation:

The captain was told to fly at at 1500 ft altitude. At the rate of 3.28 ft to 1 m, this is

\dfrac{1500}{3.28}=457 metres.

Since he was at 1450 m, he thought he was above the correct altitude by

1450 - 457 = 993 m.

In feet, this is

993 * 3.28 = 3257 ft

6 0
4 years ago
What is the most important safety rule in science? A. Never work with chemicals. B. Always use unbreakable glassware. C. Always
sasho [114]
Well based on the three I would think it's C
4 0
3 years ago
What is meant by specific latent heat of fussion​
Lunna [17]

Answer:

Boiling point of water is 100 deg C. It takes about 540 cal to change 1 g of water at the boiling point to 1 g of steam at 100 deg C.

5 0
2 years ago
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°

5 0
3 years ago
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