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Kitty [74]
2 years ago
15

If an astronaut can jump straight up to a height of 0.6 m on earth, how high could he jump on the moon?

Physics
1 answer:
Zinaida [17]2 years ago
4 0

Answer:

Explanation:

Given

Person on earth can jump to a height s=0.6\ m

initial velocity is u

using

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

final velocity is zero

s=\frac{u^2}{2g}

0.6=\frac{u^2}{2g}-----1

On moon surface acceleration due to gravity is \frac{1}{6}[/tex] th of earth gravity

so height attained is given by

h=\frac{u^2\times 6}{2\cdot g}-----2

divide 1 and 2 we get

h=6\times 0.6=3.6\ m                    

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In Ch. 1.6, the authors point out that interstellar space is not actually as empty as it seems. There is actually a lot of diffu
wel

Answer:

very small solid particles called interstellar dust.

Explanation:

In the space between the stars there is gas and dust, which represent at least 20% of the mass of our galaxy. In the Milky Way it is considered that there is a gas density of approximately 0.2 to 0.5 atoms / cm3 in the surroundings of the Sun; with respect to the dust an average of 1 g / cm3 is estimated.

Gas is about atoms and molecules, mainly hydrogen; In order of abundance, helium, carbon, oxygen, nitrogen and iron follow. On the other hand, the dust is tiny particles, generally smaller than 10 microns; the dust does not shine and therefore it is only distinguished when it is projected on bright regions (nebulae or clusters).

Interstellar matter is mainly concentrated towards the plane of the galaxy, in the strip corresponding to the Milky Way; there you can see bright nebulas of diffuse character called nebulas. These nebulae are classified according to three types: (a) bright or emission nebulae, (b) reflection nebulae and (c) planetary nebulae.

Hydrogen appears both ionized and neutral; The bright nebulae are composed of ionized hydrogen and other ionized elements. Non-ionized (neutral) hydrogen is found in the spiral arms of the Milky Way and can be detected through radio waves.

6 0
3 years ago
Calculate the AMA of an access ramp if it takes 255 N of force to push a person in a wheelchair having a combined weight of 764
Andreyy89
Actual Mechanical Advantage(AMA) = Weight / Force
Here, Weight = 764 N
Force = 255 N

Substitute the values in to the expression, 
AMA = 764 / 255
AMA = 2.99

After rounding-off to the nearest tenth value, it would be 3

Finally, option C would be your answer.

Hope this helps!
7 0
2 years ago
A car is on a circular off ramp of an interstate and is traveling at exactly 25 mph around the curve. Does the car have velocity
netineya [11]

Answer:

The car has velocity and acceleration but is not decelerating

Explanation:

Since the car is traveling at 25 mph around the curve, it has a tangential velocity. This tangential velocity is constantly changing in direction (so the car could adapt to the curve and not moving forward in a straight line), there should be a centripetal acceleration in play here. This acceleration does not slow down the car so it's not decelerating.

6 0
2 years ago
Air escaping out from an air hose at a gas station always feels cool. Why? ​
kati45 [8]

Answer:

The air is contained at a high pressure in the tube. When it escapes from a small orifice, it suddenly expands. A large amount of its heat is absorbed in the process of expansion resulting in considerable fall in its temperature. This is why the escaping air feels cold.

8 0
1 year ago
A glass flask whose volume is 1000 cm^3 at a temperature of 1.00°C is completely filled with mercury at the same temperature. W
Marat540 [252]

Answer:

the coefficient of volume expansion of the glass is \mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

Explanation:

Given that:

Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³

temperature of the glass flask and mercury= 1.00° C

After heat is applied ; the final temperature = 52.00° C

Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C

Volume of the mercury overflow = 8.50 cm^3 = 8.50 ×  10⁻⁶ m³

the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K

The increase in the volume of the mercury =  10⁻³ m³ ×  51.00 × 1.80 × 10⁻⁴

The increase in the volume of the mercury = 9.18*10^{-6} \ m^3

Increase in volume of the glass =  10⁻³ × 51.00 × \beta _{glass}

Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask

the mercury overflow = (9.18*10^{-6}  -  51.00* \beta_{glass}*10^{-3})\ m^3

8.50*10^{-6} = (9.18*10^{-6}  -51.00* \beta_{glass}* 10^{-3} )\ m^3

8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3

-6.8*10^{-7} =  ( -51.00* \beta_{glass}* 10^{-3} )\ m^3

6.8*10^{-7} =  ( 51.00* \beta_{glass}* 10^{-3} )\ m^3

\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}=  ( \beta_{glass} )

\mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

Thus; the coefficient of volume expansion of the glass is \mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

5 0
2 years ago
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