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2 years ago

A spring is stretched 6 m when a mass of 20kg is hung on it. Calculate the spring constant of this spring

Physics

1 answer:

tresset_1 [31]2 years ago

8 0

The spring constant of this spring is 9810 N/m.

<h3>What is spring constant?</h3>

Hooke's law states that the force (F) needed to extend a spring is proportional to the distance from the equilibrium position of spring.

F =kx

where k is the constant of proportionality called spring constant.

As, the mass attached with the spring is hung, then the force is equal to the weight (W).

F = W

So, kx =mg

Substituting W = mg into equation (ii) above, we have

F = mg

Substituting F = mg into equation (i) above, we have

mg = kx

Substitute, extension x = 2cm = 0.02m, mass m= 40g = 0.04kg, g is the acceleration due to gravity =9.81m/s²

Substituting these values into equation, we have

k = 20 x 9.81 /0.02

k = 9810 N/m

Therefore, the spring constant is k = 9810 N/m.

Learn more about spring constant.

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Answer:

Magnetic field can be used to produce current, infact a changing magnetic field can produce current.

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I =dPhi/dt/R

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I = AcosФ/R .dB /dt

But it isn't reasonable to say that we can create a magnetic field by having a flow of current and this can be used to make more current because the current generated due to change in magnetic field created by increase/decrease in flow of current will be in a direction such that it will counter act the change in magnetic field caused by increase/decrease in current flow.(lenz's law).

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3 0

3 years ago

Suppose that the electric field in the Earth's atmosphere is E = 8.60 101 N/C, pointing downward. Determine the electric charge

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Answer:

q = 3.87 x 10⁵ C

Explanation:

given,

Electric field, E = 8.60 x 10¹ = 86 N/C

radius of earth, R = 6371 Km = 6.371 x 10⁶ m

Coulomb constant, K = 9 x 10⁹ N · m²/C²

Charge on the earth = ?

the electric field at the point

$E =\dfrac{kq}{r^2}$

$q =\dfrac{Er^2}{k}$

inserting all the values

$q =\dfrac{86\times (6.371\times 10^6)^2}{9\times 10^{9}}$

q = 3.87 x 10⁵ C

The electric charge on the earth is equal to 3.87 x 10⁵ C

4 0

3 years ago

A 25 pF parallel-plate capacitor with an air gap between the plates is connected to a 100 V battery. A Teflon slab (dielectric c

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Answer:

The change in the charge on the positve plate when the Teflon is inserted is +2.5 nC.

Explanation:

It can be showed that the capacitance of a parallel-plate capacitor, can be expresssed as follows:

$C = \frac{\epsilon*A}{d}$

Where ε, is the dielectric constant of the material that fills the space between plates.
When this space is filled with air, ε= ε₀ = 8,85*10⁻¹² F/m.
At the same time, the capacitance of a capacitor, by definition, is as follows:

$C =\frac{Q}{V}$

If we insert a Teflon slab, in such a way that fills completely the gap between the plates, all other parameters being equal, if ε = 2*ε₀, this means that C₂ = 2* C₁. = 50 pF
As V₂=V₁ (due to the capacitor remains connected to the same battery) the charge must be the double, so Q₂ = 2* Q₁ = 5 nC.
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3 years ago

A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held hor

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Answer:

Explanation:

Given that;

horizontal circle at a rate of 2.33 revolutions per second

the magnetic field of the Earth is 0.500 gauss

the baton is 60.1 cm in length.

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we wil get the area due to rotation of radius of baton is

$\Delta A = \frac{1}{2} \Delta \theta R^2$

The formula for the induced emf is

$E = \frac{\Delta \phi}{\Delta t}$

$\phi = \texttt {magnetic flux}$

$E=\frac{\Delta (BA) }{\Delta t}$

$=B\frac{\Delta A}{\Delta t}$

B is the magnetic field strength

substitute

$\texttt {substitute}\ \frac{1}{2} \Delta \theta R^2 \ \ for \Delta A$

$E=B\frac{(\Delta \theta R^3/2)}{\Delta t} \\\\=\frac{1}{2} BR^2\omega$

The magnetic field of the earth is oriented at 14.42

$\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5$

we plug in the values in the equation above

so, the induce EMF will be

$E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega$

$=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V$