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3 years ago

6

What is the reaction force for sky diving?

1 answer:

DIA [1.3K]3 years ago

7 0

As the skydiver accelerates she gains momentum, and the air she passes through creates a resistance that pushes back up at her, increasing drag. Eventually, the force of the resisting air balances out with the force of gravity, and the skydiver stops speeding up

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What do astronomers mean when they say that the sun makes energy by hydrogen burning?

ratelena [41]

It seems that you have missed the necessary options for us to answer this question so I had to look for it. Anyway, here is the answer. What astronomers mean when <span> they say that the sun makes energy by hydrogen burning is that, the Sun is fusing hydrogen into helium and releasing energy. Hope this helps.</span>

7 0

3 years ago

2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate

adoni [48]

Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

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

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

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

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• 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.

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

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

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

4 0

2 years ago

A typical atom has a diameter of about 1.32 x 10-10 m. (a) what is this in inches (b) approximately how many atoms arethere alon

JulijaS [17]

Answer:

A ) $5.197 \times 10 ^{-9} inches$

b) There are approximately 100,000,000 atoms on the 1.32 cm lines

Explanation:

To ensure the accuracy of our answer, we will have to make sure we work in the appropriate units.

A

To convert $1.32 \times 10^{-10} m$ to inches, we can use this factor.

1 m = 39.3701 inches

Therefore $1.32 \times 10^{-10} m$ =

B

To find the number of atoms on a 1.32 cm line, we will first of all need to convert 1.32 cm to meters.

1.32 cm = 0.0132 metres (divided 1.32 by 100 to convert to metres)

Assuming all the atoms are arranged side by side, with their edges touching on the 0.0132m line, the number of atoms present will be

$0.0132/1.32\times 10^{-10}$ = 100,000,000 atoms

5 0

3 years ago

if large amount of heat are transferred to a solid, what will probably happen to the state of the solid?

Anna007 [38]

The solid can melt down to a liquid like state, and if you keep applying heat, it would eventually become a gas state.

3 0

3 years ago

Read 2 more answers

Usually the force of gravity on electrons is neglected. To see why, we can compare the force of the Earth’s gravity on an electr

miv72 [106K]

Answer:

$6.4\cdot 10^{-15} N$

Explanation:

The electric force exerted on the electron is given by:

$F=qE$

where

$q=1.6\cdot 10^{-19}C$ is the magnitude of the electron charge

E = 40000 V/m is the electric field

Substituting,

$F=(1.6\cdot 10^{-19} C)(40000 V/m)=6.4\cdot 10^{-15} N$

By comparison, the gravitational force exerted on the electron is:

$F=mg$

where

$m=9.10939\cdot 10^{-31} kg$ is the mass of the electron

g = 9.8 m/s^2 is the acceleration due to gravity

Substituting,

$F=(9.10939\cdot 10^{-31} kg)(9.8 m/s^2)=8.93\cdot 10^{-30}N$

6 0

4 years ago