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3 years ago

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A box slides with uniform acceleration up an incline. The box has an initial speed of 9.0 m/s and rises vertically 2.60 m before

coming to rest. If the angle of the incline is 30°, what is the coefficient of kinetic friction between the box and the incline?

1 answer:

Tju [1.3M]3 years ago

4 0

Answer: 0.58

Explanation:

First we need to get the acceleration of the body using equation of motion

v²=u²-2as

v is the final velocity

u is the initial velocity

a is acceleration

s is the distance moved

0²=9²-2a(2.6)

-81=-5.2a

a=81/5.2

a= 15.6m/s²

Angle of inclination =30°

To get the coefficient of friction, we use the formula

Ff =nR

Ff is frictional force

n is coefficient of friction

R is normal reaction

n = Ff/R = Wsin30°/Wcos30°

n = tan30°

n = 0.58

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Answer:

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Explanation:

Given;

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Apply the principle of conservation of linear momentum for inelastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

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Answer:

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Explanation:

u = Object distance = 30 cm

v = Image distance

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When the gold cube is immersed in mercury, the tension in the string in Newtons is 3.142N.

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In hydrogen, the transition from level 2 to level 1 has a rest wavelength of 121.6 nm.1).Find the speed for a star in which this

soldier1979 [14.2K]

Answer:

1). $v = - 2960526m/s$

2). Toward us

3). $v = - 493421m/s$

4). Toward us

5). $v = 1480263m/s$

6). Away from us

7). $v = 3207236m/s$

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Star 2: $\lambda_{measured} = 121.4nm$

Star 3: $\lambda_{measured} = 122.2nm$

Star 4: $\lambda_{measured} = 122.9nm$

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Star 4:

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$v = (3x10^{8}m/s)(\frac{122.2nm-121.6nm}{121.6nm})$

$v = 1480263m/s$

<em>Case for star 4 $\lambda_{measured} = 122.9 nm$ :</em>

$v = (3x10^{8}m/s)(\frac{122.9nm-121.6nm}{121.6nm})$

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3 years ago

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