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3 years ago

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Two people pull on a horizontal spring that is attached to an immovable wall. Then, they detach it from the wall and pull on opp

osite ends of the horizontal spring. They pull just as hard in each case. In which situation, if either, does the spring stretch more?

1 answer:

zimovet [89]3 years ago

3 0

Answer:In first case

Explanation:

When two People pull the spring, let the applied force by each person be $F_o$

and spring constant of spring be k

so Total force is $2F_o$

total extension according to Hooke's law is

$x=\frac{2F_o}{k}$

When they detach the spring and apply force in opposite direction then force on either side is F so net extension is

$x'=\frac{F_o}{k}$

so in first case there will be more extension

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Answer:

Explanation:

a)

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Ff = 0.28(1600)(9.8)cos(-84)

Ff = 458.9217...

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b) ignoring the curves required at top and bottom which change the friction force significantly, especially at the bottom where centripetal acceleration will greatly increase normal forces and thus friction force.

W = Ffd

W = 458.9217(-49.4/sin(-84)

W = 22,795.6119...

W = 23 kJ

c) same assumptions as part b

The change in potential energy minus the work of friction will be kinetic energy.

KE = PE - W

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v² = 2(gh - (μgcosθ)(h/sinθ))

v = √(2gh(1 - μcotθ))

v = √(2(9.8)(49.4)(1 - 0.28cot84))

v = 30.6552...

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2 years ago

The circumference of a sphere was measured to be

professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

$\phi = 76cm$

$Error (dr) = 0.5cm$

The radius then would be

$\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm$

And

$\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr$

PART A ) For the Surface Area we have that,

$A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}$

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

$\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)$

Maximum error:

$dA = 4(\frac{38}{\pi})(0.5)$

$dA = \frac{76}{\pi}cm^2$

The relative error is that between the value of the Area and the maximum error, therefore:

$\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}$

$\frac{dA}{A} = 0.01315 = 1.31\%$

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

$V = \frac{4}{3} \pi r^3$

$V = \frac{4}{3} \pi (\frac{38}{\pi})^3$

$V = \frac{219488}{3\pi^2}$

Therefore the Maximum Error would be,

$\frac{dV}{dr} = \frac{4}{3} 3\pi r^2$

$dV = 2r^2 (2\pi dr)$

$dV = 4r^2 (\pi dr)$

Replacing the value for the radius

$dV = 4(\frac{38}{\pi})^2(0.5)$

$dV = \frac{2888}{\pi^2} cm^3$

And the relative Error

$\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }$

$\frac{dV}{V} = 0.03947$

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2 years ago

150c passes through cell in 30 seconds. cell has a potential difference of 12v. what is current in the circuit

zzz [600]

The current in the circuit is 5 A

Explanation:

The intensity of current is given by the equation:

$I=\frac{q}{t}$

where

I is the current

q is the amount of charge passing through a given point of the circuit in a time interval of t

For the cell in this problem, we have

q = 150 C is the charge

t = 30 s is the time interval

Substituting into the equation, we f ind

$I=\frac{150}{30}=5 A$

Learn more about current:

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3 years ago

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jekas [21]

There are two atoms of potassium bonded to one atom of sulfur.

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natita [175]

Answer:

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Explanation:

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The angle of the recoiling electron with respect to the incident beam is determine by the relation :

$\cot\phi = (1+\frac{hf}{m_{e}c^{2} })\tan\frac{\theta }{2}$ ....(1)

Here ∅ is angle of recoil electron, θ is the scattered angle, h is Planck's constant, $m_{e}$ is mass of electron, c is speed of light and f is the frequency of the x-ray photon.

We know that, f = c/λ ......(2)

Here λ is wavelength of x-ray photon.

Rearrange equation (1) with the help of equation (1) in terms of λ .

$\cot\phi = (1+\frac{h}{m_{e}c\lambda })\tan\frac{\theta }{2}$

Substitute 6.6 x 10⁻³⁴ m² kg s⁻¹ for h, 9.1 x 10⁻³¹ kg for $m_{e}$ , 3 x 10⁸ m/s for c, 0.500 x 10⁻⁹ m for λ and 134° for θ in the above equation.

$\cot\phi = (1+\frac{6.6\times10^{-34} }{9.1\times10^{-31}\times3\times10^{8}\times0.5\times10^{-9} })\tan\frac{134 }{2}$

$\cot\phi=2.37$

$\phi$ = 22.90°

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