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Dimas [21]
1 year ago
10

Astronauts land on another planet and measure the density of the atmosphere on the planet surface. They measure the mass of a 50

0 cm3 conical flask plus stopper as 457.23 g. after removing the air, the mass is 456.43 g (1 m3 = 1000 litres). What is the best estimate of the density of the air?
A) 0.0000016 kg/m3
B) 0.0016 kg/m3
C) 0.16 kg/m3
D) 1.6 Kg/m3
Physics
1 answer:
Lana71 [14]1 year ago
7 0

1.6 kg/m^3 is the best estimate of the density of the air on the planet.

Given:

The mass of the conical flask with stopper is 457.23 grams and the volume is 500cm^3.

Mass of conical flask and a stopper after removing the air is 456.43 g

To find:

The density of the air on the planet.

Solution;

Mass of the conical flask and stopper with air on the planet= 457.23 g

Mass of conical flask with a stopper and without air on the planet =  456.43 g

Mass of the air in the conical flask on the planet =m

m = 457.23 g-456.43 g=0.8 g\\\\1 g = 0.001 kg\\\\m =0.8 g =0.8\times 0.001 kg=0.0008 kg

The volume of the conical flask = 500 cm^3

The volume of the air in the conical flask = V = 500cm^3

1 cm^3=10^{-6} m^3\\\\V= 500cm^3= 500\times 10^{-6}m^3=0.0005 m^3

The density of the air on the planet = d

d=\frac{m}{V}\\\\d=\frac{0.0008 kg}{0.0005 m^3}\\\\=1.6 kg/m^3

1.6 kg/m^3 is the best estimate of the density of the air on the planet.

Learn more about density here:

brainly.com/question/952755?referrer=searchResults

brainly.com/question/14373997?referrer=searchResults

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Vikentia [17]

Answer:

option B

Explanation:

given,

Force exerted by the hydraulic jack piston = F₁ = 250 N

diameter of piston, d₁ = 0.02 m

                                r₁ = 0.01 m

diameter of second piston,  d₂ = 0.15 m

                                r₂ = 0.075 m

mass of the jack to lift = ?

now,

    \dfrac{F_1}{A_1} =\dfrac{F_2}{A_2}

    \dfrac{250}{\pi r_1^2} =\dfrac{F_2}{\pi r_2^2}

    \dfrac{250}{0.01^2} =\dfrac{F_2}{0.075^2}

    F_2= \dfrac{250}{0.01^2}\times {0.075^2}

               F₂ = 14062.5 N

F = m g

m = \dfrac{F}{g}

m = \dfrac{14062.5}{9.8}

m = 1435 Kg

hence, the correct answer is option B

5 0
2 years ago
When numbers are very small or very large, it is convenient to either express the value in scientific notation and/or by using a
Oxana [17]

Answer:

5 mg, 5\cdot 10^{-3}g

Explanation:

First of all, let's rewrite the mass in grams using scientific notation.

we have:

m = 0.005 g

To rewrite it in scientific notation, we must count by how many digits we have to move the dot on the right - in this case three. So in scientific notation is

m=5\cdot 10^{-3}g

If  we want to convert into milligrams, we must remind that

1 g = 1000 mg

So we can use the proportion

1 g : 1000 mg = 0.005 g : x

and we find

x=\frac{(1000 mg)(0.005 g)}{1 g}=5 mg

4 0
2 years ago
Calculate the volume of 19 kilograms of petrol if the density of petrol is 800 kg/m?​
earnstyle [38]

Answer:

i hope this will help you :)

Explanation:

mass=19kg

density=800kg/m³

volume=?

as we know that

density=mass/volume

density×volume=mass

volume=mass/density

putting the values

volume=19kg/800kg/m³

so volume=0.02375≈0.02m³

6 0
2 years ago
one horsepower is a unit of power equal to 746w. how much energy can a 150-horsepower engine transform in 10.0s?
Dafna1 [17]

1 watt = 1 joule/second

1 horsepower = 746 watts = 746 joule/second

   (150 horsepower) x (746 watt/HP) x (1 joule/sec  /  watt) x (10 sec)

=  (150 x 746 x 1 x 10)  joule  =  1,119,000 joules .   
if correct plz mark brainly
8 0
2 years ago
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tekilochka [14]

Answer:

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Explanation:

Efficiency of a machine is defined as the ratio of useful energy to that of the energy consumed by the machine.

Here, efficiency is given as 87% and the energy consumed by the computer is 375 kWh.

Efficiency, \eta=\frac{\textrm{Useful energy}}{\textrm{Energy consumed}}

Plug in the values of \eta=0.87 and 375 kWh for energy consumed. Solve for useful energy. This gives,

Efficiency, \eta=\frac{\textrm{Useful energy}}{\textrm{Energy consumed}}\\ 0.87=\frac{\textrm{Useful energy}}{375}\\ \textrm{Useful energy}=0.87\times 375=326.25 \textrm{ kWh}

Therefore, the useful energy provided by the computer is 326.25 kWh.

3 0
2 years ago
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