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Karo-lina-s [1.5K]
3 years ago
10

An objects motion changes study island

Physics
2 answers:
dybincka [34]3 years ago
8 0

Answer:

D

Explanation:

Blizzard [7]3 years ago
3 0

Answer:

D

Explanation:

:p

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A spherical balloon is made from a material whose mass is 3.00 kg. The thickness of the material is negligible compared to the 1
vesna_86 [32]

To solve the problem it is necessary to apply the definition of Newton's second Law and the definition of density.

Density means the relationship between volume and mass:

\rho = \frac{m}{V}

While Newton's second law expresses that force is given by

F = ma

Where,

m = mass

a= acceleration (gravity at this case)

In the case of the given data we have to,

m_b = 3Kg

r = 1.5m\\V = \frac{4}{3}\pi r^3 \\V = \frac{4}{3} \pi 1.5^3\\V = 14.13m^3

In equilibrium, the entire system is equal to zero, therefore

\sum F = 0

F_g +F_h-F_b = 0

Where,

F_g = Weight of balloon

F_h = Weight of helium gas

F_b = Bouyant force

Then we have,

mg+V\rho g -V\rho_a g = 0

\rho = \rho_0-\frac{m}{V}

Replacing the values we have that

\rho = 1.19kg/m^3 -\frac{3Kg}{14.13m^3}

\rho = 0.978kg/m^3

Now by ideal gas law we have that

PV=nRT

P\frac{\rho}{m} = nRT

P = \rho \frac{n}{m}RT

But the relation \frac{n}{m} is equal to the inverse of molar mass, that is

P = \frac{\rho}{M_0} RT

P = \frac{0.978kg/m^3}{0.04kg/mol}*8.314J/K.Mol * 305K

P = 619995.7Pa

Therefore the pressure of the helium gas assuming it is ideal is 0.61Mpa

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3 years ago
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A mercury barometer is a device that is used to measure atmospheric pressure at a location. :)
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A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 2.90 m/s2 . At 20.0s after blastoff, t
Brilliant_brown [7]

Answer:

A) 580m

B) 0 m/s

C) 9.8m/s^2

D) downward

E) 10.87s

F) 106.62 m/s

Explanation:

A) The distance traveled by the rocket is calculated by using the following expression:

y=\frac{1}{2}at^2

a: acceleration of the rocket = 2.90 m/s^2

t: time of the flight = 20.0 s

y=\frac{1}{2}(2.90\frac{m}{s^2})(20.0s)^2=580m

B) In the highest point the rocket has a velocity with magnitude zero v = 0m/s because there the rocket stops.

C) The engines of the rocket suddenly fails in the highest point. There, the acceleration of the rocket is due to the gravitational force, that is 9.8 m/s^2

D) The acceleration points downward

E) The time the rocket takes to return to the ground is given by:

t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2(580m)}{9.8m/s^2}}=10.87s

10.87 seconds

F) The velocity just before the rocket arrives to the ground is:

v=\sqrt{2gy}=\sqrt{2(9.8m/s ^2)(580m)}=106.62\frac{m}{s}

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