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love history [14]

4 years ago

12

Suppose that while moving into an apartment you move a refrigerator into place by sliding it across the ﬂoor. The refrigerator w

eighs 895 N and the coefﬁcient of static friction between the ﬂoor and the refrigerator is 0.400. What is the least force you could apply to the refrigerator to cause it to move?

1 answer:

sweet-ann [11.9K]4 years ago

4 0

Answer:

358 N

Explanation:

$F=\mu N$ where F is the force, $\mu$ is the coefficient of static friction between the ﬂoor and the refrigerator and N is the weight

Normally, N=mg hence $F=\mu mg$ where m is the mass of object and g is the acceleration due to gravity

In this case, N is given as 895 N and the coefﬁcient of static friction between the ﬂoor and the refrigerator is 0.400 hence substituting them in the formula we obtain

$F=\mu N= 0.4\times 895 N= 358 N$

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The question is incomplete. Here is the complete question.

Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and then injected into a region of constant magnetic field with a field strength of 0.65T.

part (a): What is the potential difference, in volts, required in the first part of the experiment to accelerate electrons to a speed of 6.2 x 10⁷m/s?

part (b): Find the radius of curvature, in meters, of the path of a proton accelerated trhough this same potential after the proton crosses into the region with the magnetic field.

part (c) what is the ratio of the radii of curvature for a proton and an electron traveling through this apparatus?

Answer: (a) V = - 109.44 x 10² V

(b) $r_{p}=$ 9.95 x 10⁻¹ m

(c) ratio = 1800

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$\Delta V=\frac{pe}{q}$

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and its unit is joule/coulomb of Volts (V).

To determine potential difference required to accelerate a particle, we have to use the principle that the total energy of a system is conserved and one transforms into the other.

In this case, potential energy is transformed in kinetic energy:

pe = V.q

ke = $\frac{1}{2}m.v^{2}$

so

$V.q=\frac{1}{2} m.v^{2}$

$V=\frac{m.v^{2}}{2q}$

Calculating:

$V=\frac{9.11.10^{-31}(6.2.10^{7})^{2}}{2(-1.6.10^{-19})}$

V = $-109.44$ x 10²V

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Velocity and magnetic force are always perpendicular to each other. Because of that, there is no work on the particle and so, kinetic energy and speed are constant. Since magnetic force supplies centripetal force:

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$qvB=\frac{mv^{2}}{r}$

$r=\frac{mv}{qB}$

The radius of the curvature, for a proton, will be:

$r=\frac{1.67.10^{-27}.6.2.10^{7}}{1.6.10^{-19}.0.65}$

r = 9.95 x 10⁻¹m

The raius of curvature, when it is a proton, is 0.995m.

(c) Radius of curvature, if it was a electron:

$r=\frac{9.11.10^{-31}.6.2.10^{7}}{1.6.10^{-19}.0.65}$

r = 54.33 x 10⁻⁵m

ratio = $\frac{9.95.10^{-1}}{54.33.10^{-5}}$

ratio = 1800

Ratio of radii of curvature is 1800, meaning curvature created when it is a proton is 1800 times bigger than when it is a electron.

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