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2 years ago

10

A force of 6.0 N gives a 2.0 kg block an acceleration of 3.0

1 answer:

Semenov [28]2 years ago

7 0

Explanation:

The Net Force of the object can be written by:

Fnet = ma

where m is the mass of the object in <em>kg</em>

a is the acceleration of the object in <em>m/s^2</em>

Hence by applying the formula we get:

Fnet = (2.0)(3.0)

= 6N

We also know that Net force is also the sum of all forces acting on an object. In this case Friction and the Pushing Force is acting on the object. Hence we can write that:

Fnet = Pushing Force + (-Friction)

6N = 6N - Friction

Friction = 0N

Hence the<u> </u><u>f</u><u>orce of friction is 0N.</u>

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A heat pump has a coefficient of performance that is 60% of the Carnot heat pump coefficient of performance. The heat pump is us

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2 years ago

The circumference of a sphere was measured to be

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$\phi = 76cm$

$Error (dr) = 0.5cm$

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:

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

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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

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

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

$\frac{dV}{V} = 3.947\%$

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