1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
docker41 [41]
3 years ago
10

Explain how the pressure at the bottom of a container depends on the container shape and the fluid height

Physics
1 answer:
3241004551 [841]3 years ago
3 0

The pressure at the bottom of a column of fluid in a container
depends only on the depth of the fluid, not on the shape of the
container.  The pressure is simply the result of the weight of the
fluid resting on the bottom.

You might be interested in
What is a heterogeneous mixture and what is a homogeneous mixture?
yuradex [85]
A heterogeneous mixture is a mixture that contains two or more distinct substances that you can see. You can see the different part if a heterogeneous mixture. An example of this is a salad. You can see all of the parts. A homogeneous mixture is a mixture that is uniform and you cannot see the different parts. It is still a mixture though. An example of that would be salt water. The water and salt are not chemically combined but you cannot see the salt AND water. It is just one solution.
5 0
3 years ago
Read 2 more answers
A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.03 m/s on the horizontal section of a
seraphim [82]

Answer:

2.38 m/s, 4.31 m/s, lower

Explanation:

a)

Initial energy = final energy

½ m v₀² + ½ I ω₀² = mgh + ½ m v₁² + ½ I ω₁²

Since the ball is rolling without slipping, ω = v / r.

For a hollow sphere, I = ⅔ m r².

½ m v₀² + ½ (⅔ m r²) (v₀ / r)² = mgh + ½ m v₁² + ½ (⅔ m r²) (v₁ / r)²

½ m v₀² + ⅓ m v₀² = mgh + ½ m v₁² + ⅓ m v₁²

⅚ m v₀² = mgh + ⅚ m v₁²

⅚ v₀² = gh + ⅚ v₁²

v₀² = 1.2gh + v₁²

v₁ = √(v₀² − 1.2gh)

Given v₀ = 4.03 m/s, g = 9.80 m/s, h = 0.900 m:

v₁ = √((4.03)² − 1.2 (9.80) (0.900))

v₁ ≈ 2.38 m/s

At the top of the loop, the sum of the forces in the radial direction is:

∑F = ma

W + N = m v² / R

N = m v² / R - mg

N = m (v² / R - g)

Given v = 2.38 m/s, R = 0.450 m, and g = 9.80 m/s²:

N = m ((2.38)² / 0.450 - 9.80)

N = 2.77m

N ≥ 0, so the ball stays on the track.

b)

Initial energy = final energy

Borrowing from part a):

v₂ = √(v₀² − 1.2gh)

This time, h = -0.200 m:

v₂ = √((4.03)² − 1.2 (9.80) (-0.200))

v₂ ≈ 4.31 m/s

c)

Without the rotational energy:

½ m v₀² = mgh + ½ m v₁²

½ v₀² = gh + ½ v₁²

v₀² = 2gh + v₁²

v₁ = √(v₀² - 2gh)

This is less than v₁ we calculated earlier.

6 0
3 years ago
Q7:<br> A 4 kg toy is lifted off the ground and falls at 3 m/s. What is the toy's energy?
alexandr1967 [171]

Answer:

The toy's energy is 18 J.

Explanation:

We have, a 4 kg toy is lifted off the ground and falls at 3 m/s. It is required to find toy's energy.

The toy will have kinetic energy due to its motion. The energy is given by :

E=\dfrac{1}{2}mv^2\\\\E=\dfrac{1}{2}\times 4\times 3^2\\\\E=18\ J

So, the toy's energy is 18 J.

7 0
3 years ago
Topic Gravitational force amd firld strength.. help me please
I am Lyosha [343]

The gravitational force between <em>m₁</em> and <em>m₂</em> has magnitude

F_{1,2} = \dfrac{Gm_1m_2}{x^2}

while the gravitational force between <em>m₁</em> and <em>m₃</em> has magnitude

F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}

where <em>x</em> is measured in m.

The mass <em>m₁</em> is attracted to <em>m₂</em> in one direction, and attracted to <em>m₃</em> in the opposite direction such that <em>m₁</em> in equilibrium. So by Newton's second law, we have

F_{1,2} - F_{1,3} = 0

Solve for <em>x</em> :

\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

x \approx 6.74\,\mathrm m

5 0
3 years ago
How do units of work and power compare
timurjin [86]

Answer:

Jule

Explanation:

They each have the same units which is jule

3 0
3 years ago
Other questions:
  • In this type of bond, electrons are lost or gained by atoms, and the atoms are held together by electrical attraction.
    14·1 answer
  • How are electrical signals transmitted over long distances?
    6·1 answer
  • PLEASE ANSWER!!!!!
    13·1 answer
  • A truck moves 60 kilometers east from point A to point
    13·2 answers
  • Hello plz help meeeee plz I need helppp if u know it plz help
    9·1 answer
  • Explain the mode of operation of x-ray​
    13·2 answers
  • I have a voltage source of 12V but a light that only burns at 5V. The lamp works on 18 mA. Calculate the resistance that you EXT
    15·1 answer
  • Can someone pls help i’m not good at pysics
    10·2 answers
  • WILL GIVE BRAINLIEST PLEASE ANSWER ALL QUESTIONS
    10·1 answer
  • D. A bargain hunter purchases a "gold" crown at a flea market. After she gets home, she hangs it from a scale and finds its weig
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!