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klemol [59]
3 years ago
6

1.Hold one end of a meter stick down firmly on a table so that 20 centimeters of the meter stick extends past the edge of the ta

ble. Pluck the end of the meter stick that extends past the table to produce a vibration and a sound as shown. Observe the vibration and sound of the meter stick.
2: Do the same thing but at 40 centimeters this time. Observe the vibration and sound.
3: This time at 60 centimeters. Observe the vibration and sound.
Physics
1 answer:
andrew-mc [135]3 years ago
6 0
I have done this assignment before so I can tell you the more length that hangs over the table, the less vibration there is due to lack of tension.
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Glycerin at a temperature of 30 degrees celcius flows at a rate of 8×10−6m3/s through a horizontal tube with a 30mm diameter. wh
MariettaO [177]

The pressure drop in pascal is 3.824*10^4 Pascals.

To find the answer, we need to know about the Poiseuille's formula.

<h3>How to find the pressure drop in pascal?</h3>
  • We have the Poiseuille's formula,

                      Q=\frac{\pi Pr^4}{8\beta l}

where, Q is the rate of flow, P is the pressure drop, r is the radius of the pipe, \beta is the coefficient of viscosity (0.95Pas-s for Glycerin) and l being the length of the tube.

  • It is given that,

                Q=8*10^{-6}m^3/s\\diameter=30mm, thus,\\r=15mm\\l=100m\\\beta =0.95

  • Thus, the pressure drop will be,

                P=\frac{8Q\beta l}{\pi r^4} =\frac{8*8*10^{-6}*0.95*100}{3.14*(15*10^{-3})^4} \\\\P=3.824*10^4Pascals.

Thus, we can conclude that, the pressure drop in pascal is 3.824*10^4 Pascals.

Learn more about the Poiseuille's formula here:

brainly.com/question/13180459

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3 0
1 year ago
If a boulder has a mass of 50 kg and a potential energy of 490 j what is the height of the boulder
Mkey [24]
Potential energy = mass x g x height.
height = potential energy/mass x g
acceleration due to gravity on earth is 9.8 m/s
filling in your variables gives us:
490/50 x 9.8 = 1 meter
6 0
4 years ago
A meteorite strikes the Moon at the speed of 11 km/s. It has KE of 60,500,000 J. What will be its mass?
Vikentia [17]
KE=1/2 mv²
60,500,000 = 1/2 m ( 11,000 m/s )²
60,500,000= 1/2 ( 11,000 )² m
60,500,000= 60,500,000 x m
m= 60,500,000/60,500,000
m= 1 kg
8 0
3 years ago
A frictionless cart attached to a spring vibrates with amplitude A. Part A Determine what fraction of the total energy (Etot) of
Komok [63]

Answer:

the fraction of the total energy (Etot) of the cart-spring system is elastic potential energy (Us) is 0.25

the fraction is kinetic energy (K) when the cart is at position x=A/2 is 0.75

Explanation:

the solution is in the attached Word file

Download docx
5 0
3 years ago
An aseroid with a mass of 8.4x10^8 kg and a planet with a mass of 6.2x10^23 kg come close to each other by a distance of 8x10^5
N76 [4]

So, the force of gravity that the asteroid and the planet have on each other approximately

\boxed{\sf{5.43 \times 10^{10} \: N}}

<h3>Introduction</h3>

Hi ! Now, I will help to discuss about the gravitational force between two objects. We already know that <u>gravitational force occurs when two or more objects interact with each other at a certain distance and generally orbit each other to their center of mass</u>. For the gravitational force between two objects, it can be calculated using the following formula :

\boxed{\sf{\bold{F = G \times \frac{m_1 \times m_2}{r^2}}}}

With the following condition :

  • F = gravitational force (N)
  • G = gravity constant ≈ \sf{6.67 \times 10^{-11}} N.m²/kg²
  • \sf{m_1} = mass of the first object (kg)
  • \sf{m_2} = mass of the second object (kg)
  • r = distance between two objects (m)

<h3>Problem Solving</h3>

We know that :

  • G = gravity constant ≈ \sf{6.67 \times 10^{-11}} N.m²/kg²
  • \sf{m_1} = mass of the first object = \sf{8.4 \times 10^8} kg.
  • \sf{m_2} = mass of the second object = \sf{6.2 \times 10^{23}} kg.
  • r = distance between two objects = \sf{8 \times 10^5}

What was asked :

  • F = gravitational force = ... N

Step by step :

\sf{F = G \times \frac{m_1 \times m_2}{r^2}}

\sf{F = 6.67 \times 10^{-11} \times \frac{8.4 \cdot 10^8 \times 6.2 \cdot 10^{23}}{(8 \times 10^5)^2}}

\sf{F = \frac{347.374 \times 10^{-11 + 8 + 23}}{64 \times 10^10}}

\sf{F \approx  5.43 \times 10^{20 - 10}}

\boxed{\sf{F \approx 5.43 \times 10^{10} \: N}}

<h3>Conclusion</h3>

So, the force of gravity that the asteroid and the planet have on each other approximately

\boxed{\sf{5.43 \times 10^{10} \: N}}

<h3>See More</h3>

Gravity is a thing has depends on ... brainly.com/question/26485200

7 0
2 years ago
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