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Shalnov [3]
3 years ago
10

A spring had a spring constant of 48N/m. The end of the spring hangs 8m above the ground. How much weight can be placed on the s

pring so that the end of the spring is 2m above the ground

Physics
1 answer:
Oksi-84 [34.3K]3 years ago
7 0

Answer:

28.8kg

Explanation:

h1=8m

h2=2m

h2-h1=6m

k=48 N/m

g=10m/(s^2)

w=F

10m=6*48

m=28.8kg

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The Reynolds number, rho VD/mu, is a very important parameter in fluid mechanics. Verify that the Reynolds number is dimensionle
Agata [3.3K]

Answer:

Re = 1 10⁴

Explanation:

Reynolds number is

         Re = ρ v D /μ

The units of each term are

       ρ = [kg / m³]

       v = [m / s]

      D = [m]

      μ = [Pa s]

The pressure

      Pa = [N / m²] = [Kg m / s²] 1 / [m²] = [kg / m s²]

      μ = [Pa s] = [kg / m s²] [s] = [kg / m s]

We substitute the units in the equation

      Re = [kg / m³] [m / s] [m] / [kg / m s]

      Re = [kg / m s] / [m s / kg]

      RE = [ ]

Reynolds number is a scalar

Let's evaluate for the given point

Where the data for methane are:

viscosity       μ = 11.2 10⁻⁶ Pa s

the density  ρ = 0.656 kg / m³

       D = 2 in (2.54 10⁻² m / 1 in) = 5.08 10⁻² m

       Re = 0.656 4 2 5.08 10⁻² /11.2 10⁻⁶

       Re = 1.19 10⁴

4 0
3 years ago
Anne releases a stone from a height of 2 meters. She measures the kinetic energy of the stone at 9.8 joules at the exact point i
insens350 [35]
A. 0.5kg

To get this answer you need to follow the equation of KE=0.5*mv^2 
But we don't have the m part in the equation. So just plug in the numbers to see which works best, though I can tell you before we do that the answer would be a. 

As you may know, gravity, is a force of 9.8 m/s. And we want to get 9.8 Joules. So if we take a half a kg stone, release it at one meter, we get half of the normal gravity pull, 4.90 Joules. That means if we take half a kg stone and drop it at a doubled height, we get 9.8 Joules.

That is also to say that if we have a 1kg stone and drop it at one meter you will get the normal pull of gravity in Joules, 9.8J. 

Be careful though, this does not mean if you drop a 1kg stone and a .5 kg stone the 1kg will hit first. This simply means that the 1kg stone will have twice the Joules that the .5kg stone has.
7 0
3 years ago
Read 2 more answers
A 6.8 KG object moves with a velocity of 8 M/S what’s it's kinetic energy?
FromTheMoon [43]

Answer:

Explanation:

KE = ½mv² = ½(6.8)8² = 217.6 J

round as appropriate because that result is way too much precision for the inputs provided. Arguably should be 200 J based on the single significant digit of the velocity.

8 0
3 years ago
A woman with a mass of 52.0 kg is standing on the rim of a large disk that is rotating at an angular velocity of 0.470 rev/s abo
Charra [1.4K]

Explanation:

It is given that,

Mass of the woman, m₁ = 52 kg

Angular velocity, \omega=0.47\ rev/s=2.95\ rad/s

Mass of disk, m₂ = 118 kg

Radius of the disk, r = 3.9 m

The moment of inertia of woman which is standing at the rim of a large disk is :

I={m_1r^2}

I={52\times 3.9^2}

I₁ = 790.92 kg-m²

The moment of inertia of of the disk about an axis through its center is given by :

I_2=\dfrac{m_2r^2}{2}

I_2=\dfrac{118\times (3.9)^2}{2}

I₂ =897.39 kg-m²

Total moment of inertia of the system is given by :

I=I_1+I_2

I=790.92+897.39

I = 1688.31 kg-m²

The angular momentum of the system is :

L=I\times \omega

L=1688.31 \times 2.95

L=4980.5\ kg-m^2/s

So, the total angular momentum of the system is 4980.5 kg-m²/s. Hence, this is the required solution.

8 0
3 years ago
A bolt falls off an airplane high above the ground. How far does the bolt have to fall before its speed reaches 100m/s (about 20
omeli [17]
We can use the equation vf (the final velocity) =vi (the initial velocity) +at (aceleration times time)

We know the final velocity 100m/s, the initial velocity 0, and the acceleration (gravity) 9.8m/s^2. So, 100=0+9.8t. t=100/9.8
3 0
3 years ago
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