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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 solenoid 91.0 cm long has a radius of 1.50 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magni
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The magnitude of the magnetic field inside the solenoid is 6.46 \times 10^{-3} \ T.

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The magnitude of the magnetic field inside the solenoid is calculated as;

B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\

where;

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A beam of monochromatic light with a wavelength of 400 nm in air travels into water. what is the wavelength of the light in wate
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The relationship between frequency f and wavelength \lambda of a wave is given by:
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where v is the speed of the wave in the medium. The frequency of the light does not change when it moves from one medium to the other one, so we can compute the ratio between the wavelength of the light in water \lambda_w to that in air \lambda as
\frac{\lambda_w}{\lambda}= \frac{ \frac{v}{f} }{ \frac{c}{f} } = \frac{v}{c}
where v is the speed of light in water and c is the speed of light in air. Re-arranging this formula and by using \lambda=400 nm, we find
\lambda_w = \lambda \frac{v}{c}=(400 nm) \frac{2.26 \cdot 10^8 m/s}{3 \cdot 10^8 m/s}=301 nm
which is the wavelength of light in water.
5 0
2 years ago
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