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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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Suppose that a sound has initial intensity β1 measured in decibels. This sound now increases in intensity by a factor f. What is
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comparing β2 and β1, it is said that β2 is increased by a factor of f.

for each factor of f, there is a 10*f dB increase.

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A child on a 2.4 kg scooter at rest throws a 2.2 kg ball. The ball is given a speed of 3.1 m/s and the child and scooter move in
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Answer:

The child's mass is 14.133 kg

Explanation:

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Where:

m₁ = Mass of the child

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v₃ = Final velocity of the child and scooter = 0.45 m/s

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Plugging the values gives;

(m₁ + 2.4)× 0.45 = 2.4 × 3.1

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3 0
2 years ago
1. California sea lions can swim as fast as 40.0 km/h. Suppose a sea lion begins to chase a fish at this speed when the fish is
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#1

In order to chase the fish the distance traveled by sea lion in time t must be equal to the distance of sea lion from the fish and distance traveled by fish in the same time.

So here we can say let say sea lion chase the fish in time "t"

then here we have

d_1 = d_2 + L

here

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d2 = distance covered by fish in the same time t

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# 2

velocity of truck on road = 25 m/s along North

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we can write the relative velocity as

v_d - v_t = 1.43 \hat j + 1 \hat i

v_d = v_t + (1.43 \hat j + 1 \hat i)

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8 0
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