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maxonik [38]
2 years ago
10

A spring with a spring constant of 400 n / M has a mass hung on it so it stretches 8 cm. calculate how much mass the spring is s

upporting.
Physics
1 answer:
I am Lyosha [343]2 years ago
5 0

Answer:

3.3kg

Explanation:

Given parameters:

Spring constant = 400N/m

Extension  = 8cm  = 0.08m

Unknown:

Amount of mass the spring is supporting  = ?

Solution:

To solve this problem:

     F  = kE

F is the force

k is the spring constant

E is the extension

  So;

            F  = 400 x 0.08  = 32N

Mass;

       Force  = mass x acceleration due to gravity

          32  = mass x 9.8

  Mass = 3.3kg

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
2 years ago
In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of rad
Crank

Answer:

a)   Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1] , b) Q = 3.4 10⁻² m³ / s , c)      Q = 4.8 10⁻² m³ / s

Explanation:

We can solve this fluid problem with Bernoulli's equation.

         P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

With the two tubes they are at the same height y₁ = y₂

        P₁-P₂ = ½ ρ (v₂² - v₁²)

The flow rate is given by

         A₁ v₁ = A₂ v₂

         v₂ = v₁ A₁ / A₂

We replace

         ΔP = ½ ρ [(v₁ A₁ / A₂)² - v₁²]

         ΔP = ½ ρ v₁² [(A₁ / A₂)² -1]

Let's clear the speed

         v₁ = √ 2ΔP /ρ[(A₁ / A₂)² -1]

The expression for the flow is

           Q = A v

           Q = A₁ v₁

           Q = A₁ √ 2ΔP / rho [(A₁ / A₂)² -1]

The areas are

            A₁ = π r₁

            A₂ = π r₂

We replace

        Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1]

Let's calculate for the different pressures

      r₁ = d₁ / 2 = 1.00 / 2

      r₁ = 0.500 10⁻² m

      r₂ = 0.250 10⁻² m

b) ΔP = 6.00 kPa = 6 10³ Pa

      Q = π 0.5 10⁻² √(2 6.00 10³ / (850 (0.5² / 0.25² -1))

       Q = 1.57 10⁻² √(12 10³/2550)

        Q = 3.4 10⁻² m³ / s

c) ΔP = 12 10³ Pa

        Q = 1.57 10⁻² √(2 12 10³ / (850 3)

         Q = 4.8 10⁻² m³ / s

5 0
2 years ago
Imagine that the above hoop is a tire. the coefficient of static friction between rubber and concrete is typically at least 0.9.
Stels [109]
The hoop is attached.

Consider that the friction force is given by:
F = μ·N
   = μ·m·g·cosθ

We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·<span>g·sinθ

Therefore:
</span>-(1/2)·m·g·sinθ + m·g·sinθ = <span>μ·m·g·cosθ
</span>(1/2)·m·g·sinθ = <span>μ·m·g·cosθ
</span>μ = (1/2)·m·g·sinθ / <span>m·g·cosθ
   = </span>(1/2)·tanθ

Now, solve for θ:
θ = tan⁻¹(2·μ)
   = tan⁻¹(2·0.9)
   = 61°

Therefore, the maximum angle <span>you could ride down without worrying about skidding is 61°.</span>

5 0
2 years ago
What is the total energy q released in a single fusion reaction event for the equation given in the problem introduction? use c2
kkurt [141]

Answer:

4.3\cdot 10^{-12} J

Explanation:

The fusion reaction in this problem is

4^1_1H \rightarrow ^4_2He +2e^+

The total energy released in the fusion reaction is given by

\Delta E = c^2 \Delta m

where

c=3.0\cdot 10^8 m/s is the speed of light

\Delta m is the mass defect, which is the mass difference between the mass of the reactants and the mass of the products

For this fusion reaction we have:

m(^1_1H)=1.007825u is the mass of one nucleus of hydrogen

m(^4_2 He)=4.002603u is the mass of one nucleus of helium

So the mass defect is:

\Delta m =4m(^1_1 H)-m(^4_2 He)=4(1.007825u)-4.002603u=0.028697u

The conversion factor between atomic mass units and kilograms is

1u=1.66054\cdot 10^{-27}kg

So the mass defect is

\Delta m =(0.028697)(1.66054\cdot 10^{-27})=4.765\cdot 10^{-29}kg

And so, the energy released is:

\Delta E=(3.0\cdot 10^8)^2(4.765\cdot 10^{-29})=4.3\cdot 10^{-12} J

6 0
2 years ago
The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal cha
saveliy_v [14]

Answer:

d. The scale's resolution is too low to read the change in mass

Explanation:

If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:

F = kx

since,

w = Fd

dw = Fdx

integrating and using value of F, we get:

ΔE = (0.5)kx²

where,

ΔE = Energy added to spring

k = spring constant

x = displacement

The spring constant is typically in range of 4900 to 29400 N/m.

So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:

ΔE = (0.5)(29400 N/m)(1 m)²

ΔE = 1.47 x 10⁴ J

Now, if we convert this energy to mass from Einstein's equation, we get:

ΔE = Δmc²

Δm = ΔE/c²

Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²

<u>Δm =  4.9 x 10⁻¹³ kg</u>

As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

Thus, it can not determine such a small change. So, the correct option is:

<u>d. The scale's resolution is too low to read the change in mass</u>

8 0
2 years ago
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