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Andrei [34K]
3 years ago
7

if you poured oil and water into a beaker , which liquid would be on top and which would be on the bottom ? How would the densit

ies of the liquids compare with their positions in the beaker ? Explain
Physics
1 answer:
Effectus [21]3 years ago
7 0

Answer:

 oil floats in water due to the fact that it is less dense. since the water is more dense, it is basically heavier, and heavier items always sink, while lighter items float, the same can be said about liquids. denser liquids will sink while less dense items will float.

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A boy finds an abandoned mine shaft in the woods, and wants to know how deep the hole is. He drops in a stone, and counts 4 seco
Shalnov [3]

The shaft is 78m approximately deep. The correct option is D which is 78 meters

<h3>What are Sound Waves ?</h3>

Sound waves are longitudinal. That is, the direction of the waves is parallel to the direction of its propagation of particles.

Given that a boy finds an abandoned mine shaft in the woods, and wants to know how deep the hole is. He drops in a stone, and counts 4 seconds before he hears the "plunk" of the stone hitting the bottom of the shaft.

  • The speed of the sound V = 330m/s
  • The time it takes the sound to reach the top = t

Speed V = distance / time

330 = h/t

Make t the subject of formula

t = h/330

As the stone is dropped, initial velocity = 0, Using the formula

h = 1/2gT²

But T = 4 - t

T = 4 approximately

Substitute all the parameters

h = 1/2 × 9.8 × (4)²

h = 4.9 × (16)

h = 78.4m

Therefore, the shaft is 78m approximately deep. The correct option is D which is 78 meters

Learn more about sound wave here: brainly.com/question/16093793

#SPJ1

3 0
2 years ago
A car with mass 2.0×10 3 kg traveling east at a speed of 20 m/s collides at an intersection with a2.5×10 3 kg van traveling nort
lys-0071 [83]
When analyzing inelastic collisions, we need to consider the law of conservation of momentum, which states that the total momentum, p, of the closed system is a constant. In the case of inelastic collisions, the momentum of the combined mass after the collision is equal to the sum of the momentum of each of the initial masses.

p1+p2+...=pf

In our case we only have two masses, which makes our problem fairly simple. Lets plug in the formula for momentum; p=mv.

m1v1+m2v2=(m1+m2)vf

To find the velocity of the combined mass we simply rearrange the terms.

vf=m1v1+m2v2m1+m2

Plug in the values given in the problem.

vf=(3.0kg)(1.4m/s)+(2.0kg)(0m/s)03.0kg+2.0kg

vf=.84m/s

7 0
3 years ago
Calculate the momentum of a 147 kg tiger running south at 14.7m/s.
Ganezh [65]
P=mv

2087.4 kg meter/second
7 0
3 years ago
How would doubling the height of an object change the object's potential energy?
Kruka [31]

C. The potential energy would double.

Explanation:

The potential energy of an object is the energy at rest. it is the product of its mass, gravity and height.

   Potential energy = mgh

where m is the mass

           g is the acceleration due to gravity

            h is the height

From the above, potential energy is proportional to the height. If any change in any of the parameters will directly affect the potential energy. This implies that if height is double, the potential energy is doubled.

Learn more:

Potential energy brainly.com/question/10770261

#learnwithBrainly

6 0
3 years ago
A pendulum of mass 5.0 kg hangs in equilibrium. A frustrated student walks up to it and kicks the bob with a horizontal force of
Phoenix [80]

Answer:

The length is L = 6.206m and the angle is \theta = 37.752^o.

Explanation:

The period T of the pendulum is related to its length L by

T = 2\pi \sqrt{\dfrac{L}{g} },

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

Solving for L we get

L = \dfrac{T^2g}{4\pi^2}

putting in T =5.0s and g =9.8m/s^2 we get:

L = \dfrac{(5.0s)^2*9.8m/s^2}{4\pi^2}

\boxed{L = 6.206m.}

There are two forces acting on the pendulum: The gravitational force mg and the F = 30N student's force. Therefore, the angular displacement \theta that these forces give is

sin(\theta ) = \dfrac{F}{mg}

\theta = sin^{-1}( \dfrac{F}{mg})

putting in F =30N, m =5.0kg, and g =9.8m/s^2 we get

\theta = sin^{-1}( \dfrac{30N}{5.0kg*9.8m/s^2})

\boxed{\theta = 37.752^o.}

4 0
4 years ago
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