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zmey [24]
3 years ago
7

11. A 2.5-kg block slides down a 25o inclined plane with a constant acceleration. The block starts from rest at the top. At the

bottom, its velocity reaches 0.65 m/s. The length of the incline is 1.6m. a) What is the acceleration of the block?
b) What is the coefficient of friction between the plane and the block?
c) Does the result of either (a) or (b) depend on the mass of the block?
Physics
1 answer:
erastovalidia [21]3 years ago
5 0

Answer:

Explanation:

Given

mass of Block m=2.5 kg

velocity at bottom v=0.65 m/s

Length of Ramp L=1.6 m

inclination \theta =25^{\circ}

using v^2-u^2=2 as

where a=acceleration

s=displacement

v=final velocity

u=initial velocity

0.65^2-0=2\times a\times 1.6

a=0.132 m/s^2

(b)Block is moving downward with constant acceleration is given by

F_{net}=mg\sin \theta -\mu mg\cos \theta

a=g\sin \theta -\mu g\cos \theta

0.132=g(\sin 25-\mu \cos 25)

\mu \cdot 0.906=0.4360

\mu =\frac{0.4360}{0.906}

\mu =0.481

Result will remain same irrespective of mass

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A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with
Julli [10]

Explanation:

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(B). We need to calculate the V_{R} when f = \dfrac{1}{2f_{c}}

Using formula of  V_{R}

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Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})

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Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=4.561\ Volt

Hence, This is the required solution.

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3 years ago
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