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Masteriza [31]
4 years ago
14

Compared to the buoyant force of the atmosphere on a 1-kilogram helium-filled balloon, the buoyant force of the atmosphere on a

nearby 1-kilogram solid iron block is _____.
Physics
2 answers:
bezimeni [28]4 years ago
7 0

Answer:

Explanation:

The buoyant force depends on the volume of the body

Buoyant force = Volume immersed x density of fluid x gravity

So, the density of fluid that means air is same for both the cases, but the volume of 1 kg helium balloon is more than the density of iron, so the buoyant force on helium is more than the buoyant force on iron.

Vladimir [108]4 years ago
6 0

It all depends on the SIZE of the balloon.

If the balloon is made of really tough rubber, and it holds the helium in the same volume as the solid iron block, then the buoyant force of the atmosphere is the same for both objects.

But if the balloon is just some flimsy stuff, and it lets the helium expand to a much bigger volume than the iron block, then the buoyant force on the balloon is greater than the buoyant force on the solid iron block.

In fact, it DOESN'T MATTER what's in the balloon and what's in the block.  It doesn't matter whether either one of them is solid, liquid, or gas, and it doesn't matter whether they have the same or different mass.  

Whichever one has greater VOLUME has a greater buoyant force of atmosphere acting on it.

You might be interested in
A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f
konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

(c) The maximum speed attained by the object during its motion is 0.577 m/s

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m

(c) the maximum speed attained by the object during its motion

E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s

8 0
4 years ago
A solid sphere of radius R is made of a metallic conductor. A hollow spherical shell of the same radius R is made of the same co
Harrizon [31]

1-2) They have same surface charge density

3-4) The metallic conductor has greatest surface charge density

Explanation:

1-2)

In a conductor, the charge carriers (mainly electrons) are free to move. Therefore, as a result, they tend to move at the largest possible distance from each other, because of the repulsive force that they exert on each other.

The configuration that maximize the distance between the charge carriers for a solid sphere of metallic conductor is the one in which all the electrons are on the surface, and they are equally spaced between each other. This means that for the solid sphere of radius R, the excess charge Q will be entirely spread over the surface of the sphere.

Similarly, the excess charge Q on the hollow spherical shell (which is also made of the same conducting material) will also be spread over the surface with the charge carriers at the maximum distance from each other. Therefore, the surface charge density for both objects will be

\sigma = \frac{Q}{4\pi R^2}

where R is the radius of the two spheres.

3-4)

In this case, the surface charge density on the two objects is different.

In fact, on the metallic sphere (conducting) the surface charge density is (as explained in part 1):

\sigma = \frac{Q}{4\pi R^2}

Hoever, the second sphere is made of an insulating material. In an insulator, the charge carriers are not free to move. If the initial charge Q is spread across the all sphere (which is not hollow), this means that some of the charge will actually also be inside the sphere. So the charge deposited on the surface, Q', will be less than the total charge Q. Therefore, the surface charge density will be

\sigma' = \frac{Q'}{4\pi R^2}

And since Q' < Q, this means that \sigma', so the conducting sphere has a greatest surface charge  density.

4 0
3 years ago
The literary movement with which Wilde is most closely associated is Select one: a. romanticism b. existentialism c. Art for art
jeyben [28]

Answer:

Correct answer is ''c'' Art for art's Sake

Explanation:

Wilde wrote in the literary movement called Aestheticism during the late nineteenth century. Contrary to popular belief, Wilde did not create the literary movement, but played a role as a leader who promoted the movement. While Wilde was a college student the works of Algernon Charles Swinburne and Edgar Allan Poe influenced his own writing style. Also, the English essayist Walter Pater helped to form Wilde's humanistic aesthetics.

The philosophical foundations of Aestheticism come from Immanuel Kant. He formulated the idea of "art for art's sake". He believed that art was to be enjoyed for its own beauty regardless of social or moral concerns.

6 0
3 years ago
A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizo
blondinia [14]

Answer:

His pitching speed is 38 m/s.

Explanation:

Hi there!

Please see the attached figure for a better understanding of the problem.

The position of the ball at any time t is given by the following vector:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector of the ball at time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Let's place the origin of the frame of reference at the throwing point so that x0  and y0 = 0.

When the ball reaches the ground, its position vector will be r1 (see figure). Using the equation of the vertical component of the position vector, we can find the time at which the ball reaches the ground. At that time, the horizontal component of the position is 30 m and the vertical component is -3.0 m (see figure):

y = y0 + 1/2 · g · t²  (y0 = 0)

y = 1/2 · g · t²

-3.0 m = 1/2 · (-9.8 m/s²) · t²

-3.0 m / -4.9 m/s² = t²

t = 0.78 s

Now, knowing that at this time x = 30 m, we can find v0:

x = x0 + v0 · t  (x0 = 0)

x = v0 · t

30 m = v0 · 0.78 s

v0 = 30 m / 0.78 s

v0 = 38 m/s

His pitching speed is 38 m/s.

3 0
4 years ago
A spherically symmetric charge distribution has a charge density given by ρ = a/r , where a is constant. Find the electric field
Airida [17]
Charge dQ on a shell thickness dr is given by 

dQ = (charge density) × (surface area) × dr 

dQ = ρ(r)4πr²dr 

∫ dQ = ∫ (a/r)4πr²dr 

∫ dQ = 4πa ∫ rdr 

Q(r) = 2πar² - 2πa0² 

Q = 2πar² (= total charge bound by a spherical surface of radius r) 

Gauss's Law states: 

(Flux out of surface) = (charge bound by surface)/ε۪ 

(Surface area of sphere) × E = Q/ε۪ 

4πr²E = 2πar²/ε۪ 

<span>E = a/2ε۪


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

</span>
3 0
3 years ago
Read 2 more answers
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