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3 years ago

What is Circular Motion?

2 answers:

5 0

Circular motion is what an object has if it is moving around and around and around and around and around and around and around and around and around in a path that is a circle.

andre [41]3 years ago

3 0

It is a movement of an object along the circumference of a circle

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Imagine a particular exoplanet covered in an ocean of liquid methane. At the surface of the ocean, the acceleration of gravity i

AlladinOne [14]

Answer:

Explanation:

Atmospheric pressure = 7 x 10⁴ Pa

force on a disk-shaped region 2.00 m in radius at the surface of the ocean due to atmosphere = pressure x area

= 7 x 10⁴ x 3.14 x 2 x 2

= 87.92 x 10⁴ N

b )

weight, on this exoplanet, of a 10.0 m deep cylindrical column of methane with radius 2.00 m

Pressure x area

height x density x acceleration of gravity x π r²

= 10 x 415 x 6.2 x 3.14 x 2 x 2

=323168.8 N

c ) Pressure at a depth of 10 m

atmospheric pressure + pressure due to liquid column

= 7 x 10⁴ + 10 x 415 x 6.2 ( hρg)

= 7 x 10⁴ + 10 x 415 x 6.2

(7 + 2.57 )x 10⁴ Pa

9.57 x 10⁴ Pa

8 0

3 years ago

A large balloon of mass 210 kg is filled with helium gas until its volume is 329 m3. Assume the density of air is 1.29 kg/m3 and

Nastasia [14]

(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

There are only two forces acting on the balloon, if we neglect air resistance:

- The weight of the balloon, labelled with W, whose magnitude is

$W=mg$

where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity, and where the direction is upward

(b) 4159 N

The buoyant force is given by

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity.

In this case we have

$\rho_a = 1.29 kg/m^3$ is the air density

$V=329 m^3$ is the volume of the balloon

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

So the buoyant force is

$B=(1.29 kg/m^3)(329 m^3)(9.8 m/s^2)=4159 N$

(c) 1524 N

The mass of the helium gas inside the balloon is

$m_h=\rho_h V=(0.179 kg/m^3)(329 m^3)=59 kg$

where $\rho_h$ is the helium density; so we the total mass of the balloon+helium gas inside is

$m=m_h+m_b=59 kg+210 kg=269 kg$

So now we can find the weight of the balloon:

$W=mg=(269 kg)(9.8 m/s^2)=2635 N$

And so, the net force on the balloon is

$F=B-W=4159 N-2635 N=1524 N$

(d) The balloon will rise

Explanation: we said that there are only two forces acting on the balloon: the buoyant force, upward, and the weight, downward. Since the magnitude of the buoyant force is larger than the magnitude of the weigth, this means that the net force on the balloon points upward, so according to Newton's second law, the balloon will have an acceleration pointing upward, so it will rise.

(e) 155 kg

The maximum additional mass that the balloon can support in equilibrium can be found by requiring that the buoyant force is equal to the new weight of the balloon:

$W'=(m'+m)g=B$

where m' is the additional mass. Re-arranging the equation for m', we find

$m'=\frac{B}{g}-m=\frac{4159 N}{9.8 m/s^2}-269 kg=155 kg$

(f) The balloon and its load will accelerate upward.

If the mass of the load is less than the value calculated in the previous part (155 kg), the balloon will accelerate upward, because the buoyant force will still be larger than the weight of the balloon, so the net force will still be pointing upward.

(g) The decrease in air density as the altitude increases

As the balloon rises and goes higher, the density of the air in the atmosphere decreases. As a result, the buoyant force that pushes the balloon upward will decrease, according to the formula

$B=\rho_a V g$

So, at a certain altitude h, the buoyant force will be no longer greater than the weight of the balloon, therefore the net force will become zero and the balloon will no longer rise.

4 0

3 years ago

A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi

slega [8]

We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is
$dI =r^2*dm$
where $dm =M(b*dr)/(ab)$
Now we find the moment of inertia by integrating from $-a/2$ to $a/2$
The moment of inertia is
$I= \int\limits^{-a/2}_{a/2} {r^2*dm} = M \int\limits^{-a/2}_{a/2} r^2(b*dr)/(ab)=(M/a)(r^3/3)$ (from (-a/2) to $I=(M/3a)(a^3/8 +a^3/8)=(Ma^2)/12$ (a/2))

4 0

3 years ago

How much work must be done to stop a 1100-kg car traveling at 112 km/h?(Hint: You will need to convert the speed first.)Answer:

zimovet [89]

According to the Work-Energy Theorem, the work done on an object is equal to the change in the kinetic energy of the object:

$W=\Delta K$

Since the car ends with a kinetic energy of 0J (because it stops), then the work needed to stop the car is equal to the initial kinetic energy of the car:

$K=\frac{1}{2}mv^2$

Replace m=1100kg and v=112km/h. Write the speed in m/s. Remember that 1m/s = 3.6km/h:

$\begin{gathered} K=\frac{1}{2}(1100kg)\left(112\frac{km}{h}\times\frac{1\frac{m}{s}}{3.6\frac{km}{h}}\right)^2=532,345.679...J \\ \\ \therefore K\approx532,346J \end{gathered}$

Therefore, the answer is: 532,346 J.

5 0

1 year ago

Calculating average speed

djverab [1.8K]

Average speed =

(distance covered during some period of time)
divided by
(length of time to cover that distance).

7 0

3 years ago