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2 years ago

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!pls

Physics

1 answer:

barxatty [35]2 years ago

5 0

I think this is the solution:

1: U-1, F,-4
2: Na-6, Mo-1, O-4
3: Bi-1, O-1, C-1, I-1
4: In-9, N-1
5: N-2, H-4, S-1, C-1
6: Ge- 15, N-4
7: N-1, H-4, C-1, I-1, O-3
8: H-7, F-1
9: N-1, O-5, H-1, S-1
10: H-8
11: Nb-1, O-1, C-1, I-3
12: C-3, F-3, S-1, O-3, H-1
13: Ag-1, C-1, N-1, O-1
14: Pb-6, H-1, As-1, O-4

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adoni [48]

Answer:

 » Image distance :

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

v is image distance
u is object distance, u is 10 cm
f is focal length, f is 5 cm

${ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}$

 » Magnification :

• Let's derive this formula from the lens formula:

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

» Multiply throughout by fv

${ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}$

• But we know that, v/u is M

${ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = \frac{v}{f} }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = \frac{v}{f} - 1 }}}}$

v is image distance, v is 10 cm
f is focal length, f is 5 cm
M is magnification.

${ \tt{M = \frac{10}{5} - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}$

 » Nature of Image :

Image is magnified
Image is erect or upright
Image is inverted
Image distance is identical to object distance.

4 0

2 years ago

Please help in this question!! I will mark brainlest

LUCKY_DIMON [66]

Answer:

im not too sure about that all i know is history

4 0

3 years ago

In the diagram, the wavelength is shown by :

MAXImum [283]

Answer:

i think the anwer is C

Explanation:

6 0

3 years ago

A subway train accelerates from rest at one station at a rate of 1.30 m/s^2 for half of the distance to the next station, then d

Minchanka [31]

This problems a perfect application for this acceleration formula:

 Distance = (1/2) (acceleration) (time)² .

During the speeding-up half: 1,600 meters = (1/2) (1.3 m/s²) T²
During the slowing-down half: 1,600 meters = (1/2) (1.3 m/s²) T²

Pick either half, and divide each side by 0.65 m/s²:

 T² = (1600 m) / (0.65 m/s²)

 T = square root of (1600 / 0.65) seconds

Time for the total trip between the stations is double that time.

 T = 2 √(1600/0.65) = 99.2 seconds (rounded)

7 0

3 years ago

An 16.00-kg object is suspended by a rope at some height, h if the object has a potential energy of 900.01j what is the objects

qwelly [4]

Answer:

5.740 m

Explanation:

PE = mgh

900.0 J = (16.00 kg) (9.8 m/s²) h

h = 5.740 m

5 0

3 years ago

Read 2 more answers