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iren [92.7K]
3 years ago
14

A wooden block of mass m = 9 kg starts from rest on an inclined plane sloped at an angle θ = 30° from the horizontal. The block

is originally located 5m from the bottom of the plane. If the block, undergoing constant acceleration down the ramp, slides to the bottom in 2 seconds, what is the coefficient of the kinetic friction μk between the block and the inclined plane?
Physics
1 answer:
jenyasd209 [6]3 years ago
5 0

Answer:

Fk = 21.645N

Explanation:

Let Fb be Force of block and thus;

Fb= mg where m is mass of block and g is acceleration due to gravity

Thus Fb= 9kg x 9.81N/kg = 88.29 N

Now, the question says this force Fb rests at an inclined plane [email protected] 30° angle

Thus;

Force parallel to inclined plane = 88.29 x sin30° = 44.145 N.

Force perpendicular to the inclined plane = 88.29 x cos30 = 76.46 N

Now, when an object is falling freely, we know that

h = (1/2)at^(2)

From the question, the height is 5m and t= 2 seconds

Thus;

5 = (1/2)a(2)^(2)

2a = 5 and thus,

a = 5/2 = 2.5 m/s^(2)

Now, in inclined planes, perpendicular force - kinetic friction force = Resultant force

Thus let perpendicular be Fp and kinetic friction force be Fk and so;

Fp - Fk = F

F= ma = 9 x 2.5 = 22.5N

Thus, 44.145 - Fk = 22.5

Thus, Fk = 44.145 - 22.5 = 21.645N

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You have 2 resistors of unknown values you label Ra and Rb. You have an old battery and a multimeter you bought years ago for 7$
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Answer:

the answers, the correct one is D,   Rb₂ = 29.97 ohm

Explanation:

For this exercise we use ohm's law and the equivalent resistance ratio for series and parallel circuits.

Serial circuit

         (Ra + Rb) is = V

         (Ra + Rb) 0.111 = 10

         (Ra + Rb) = 10 / 0.111 = 90.09

parallel circuit

         \frac{1}{R} = \frac{1}{Ra} + \frac{1}{Rb}

           R = \frac{Ra \ Rb}{Ra + Rb}

          \frac{Ra \ Rb}{Ra + Rb} i_p = V

         \frac{Ra \ Rb}{Ra + Rb} 0.5 = 10

         \frac{Ra \ Rb}{Ra + Rb} = 10 / 0.5 = 20

we write and solve our system of equations

         Ra + Rb = 90.09

         \frac{Ra \ Rb}{Ra + Rb} = 20

         

we solve for Ra in the first equation

          Ra = 90.09 - Rb

           RaRb = 20 (Ra + Rb)

we substitute Ra in the second equation

           (90.09-Rb) Rb = 20 [(90.09-Rb) + Rb]

            90.09 Rb - Rb² = 20 90.09

           

            Rb² - 90.09 Rb + 1801.8 = 0

we solve the quadratic equation

            Rb = [90.09 ±\sqrt{90.09^2 - 4 \ 1801.8} ] / 2

            Rb = [90.09 ± 30.15] / 2

            Rb₁ = 60.12 ohm

            Rb₂ = 29.97 ohm

the smallest value is Rb = 30 ohm

When checking the answers, the correct one is D

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Answer:

A uniform thin rod with an axis through the center

Consider a uniform (density and shape) thin rod of mass M and length L as shown in (Figure). We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x-axis.

We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write

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