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2 years ago

6

A student wants to determine the coefficient of static friction μ between a block of wood and an adjustable inclined plane. Of t

he following, the minimum additional equipment the student needs to determine a value for μ is

1 answer:

Cloud [144]2 years ago

7 0

Answer:

A protractor to measure the angle of the inclined plane with the horizontal

Explanation:

The student needs to lift the free end of the adjustable inclined plane until the object barely starts sliding, and measure the angle at which such happens. At that point, the force of friction equals the component of the weight in the direction of the incline. That is:

$f=\mu\,* N = \mu * m g\, cos(\theta)$

and $w_{//}= m\,g\,sin(\theta)$

Then

$f = w_{//}\\\mu *\,m \,g\,cos(\theta) = m\,g\,sin(\theta)\\\mu = tan(\theta)$

and therefore, the coefficient of static friction is fully determined just by calculating the tangent of the angle that the incline forms with the horizontal.

Then the only extra instrument needed is a protractor to measure the angle.

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A thin copper rod 1.0 m long has a mass of 0.050 kg and is in a magnetic field of 0.10 t. What minimum current in the rod is nee

slamgirl [31]

Answer:

$i = 4.9 A$

Explanation:

Force on a current carrying rod due to magnetic field is given as

$F = iLB$

here we know that

$i =$ current in the rod

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$L = 1.0 m$

now magnetic force is balanced by the weight of the rod

so we will have

$iLB = mg$

$i(1.0)(0.10) = 0.05 \times 9.8$

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2 years ago

A very long line of charge with charge per unit length +8.00 μC/m is on the x-axis and its midpoint is at x = 0. A second very l

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Answer:

at y=6.29 cm the charge of the two distribution will be equal.

Explanation:

Given:

linear charge density on the x-axis, $\lambda_1=8\times 10^{-6}\ C$

linear charge density of the other charge distribution, $\lambda_2=-6\times 10^{-6}\ C$

Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.

Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.

<u>we know, the electric field due to linear charge is given as:</u>

$E=\frac{\lambda}{2\pi.r.\epsilon_0}$

where:

$\lambda=$ linear charge density

r = radial distance from the center of wire

$\epsilon_0=$ permittivity of free space

Therefore,

$E_1=E_2$

$\frac{\lambda_1}{2\pi.x.\epsilon_0}=\frac{\lambda_2}{2\pi.(0.11-x).\epsilon_0}$

$\frac{\lambda_1}{x} =\frac{\lambda_2}{0.11-x}$

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2 years ago

1. Two identical bowling balls of mass M and radius R roll side by side at speed v0 along a flat surface. Ball 1 encounters a ra

UNO [17]

Answer:

1/2

Explanation:

We need to make a couple of considerations but basically the problem is solved through the conservation of energy.

I attached a diagram for the two surfaces and begin to make the necessary considerations.

Rough Surface,

We know that force is equal to,

$F_r = mgsin\theta$

$F_r = \mu N$

$F_r = \mu mg cos\theta$

Matching the two equation we have,

$\mu N = \mu mg cos\theta$

$\mu = tan\theta$

Applying energy conservation,

$\frac{1}{2}mv^2_0+\frac{1}{2}I_w^2 = F_r*d+mgh_1$

$\frac{1}{2}mv^2_0+\frac{2}{5}mR^2\frac{V_0^2}{R^2} = F_r*d+mgh_1$

$\frac{1}{2}mv^2_0+\frac{mv_0^2}{5} = mgsin\theta \frac{h_1sin\theta}+mgh_1$

$\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_1+gh_1$

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Frictionless surface

$\frac{1}{2}mv_0^2+\frac{1}{2}I\omega^2 = mgh_2$

$\frac{1}{2}m_v^2+\frac{1}{2}\frac{2}{5}mR^2\frac{v_0^2}{R^2} =mgh_2$

$\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_2$

$h_2 = \frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})$

Given the description we apply energy conservation taking into account the inertia of a sphere. Then the relation between $h_1$ and $h_2$ is given by

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$\frac{h_1}{h_2} = \frac{1}{2}$

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A 1.00-kilogram ball is dropped from the top of a building. just before striking the ground, the ball's speed is 12.0 meters per

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