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UkoKoshka [18]
3 years ago
14

A bin is given a push across a horizontal surface. The bin has a mass m, the push gives it an initial speed of 1.60 m/s, and the

coefficient of kinetic friction between the bin and the surface is 0.150. (a) Use energy considerations to find the distance (in m) the bin moves before it stops. m (b) What If
Physics
1 answer:
emmasim [6.3K]3 years ago
5 0

Answer:

The bin moves 0.87 m before it stops.

Explanation:

If we analyze the situation and apply the law of conservation of energy to this case, we get:

Energy Dissipated through Friction = Change in Kinetic Energy of Bin (Loss)

F d = (0.5)(m)(Vi² - Vf²)

where,

F = Frictional Force = μR    

but, R = Normal Reaction = Weight of Bin = mg

Therefore, F = μmg

Hence, the equation becomes:

μmg d = (0.5)(m)(Vi² - Vf²)

μg d = (0.5)(Vi² - Vf²)

d = (0.5)(Vi² - Vf²)/μg

where,

Vf = Final Velocity = 0 m/s (Since, bin finally stops)

Vi = Initial Velocity = 1.6 m/s

μ = coefficient of kinetic friction = 0.15

g = 9.8 m/s²

d = distance moved by bin before coming to stop = ?

Therefore,

d = (0.5)[(1.6 m/s)² - (0 m/s)²]/(0.15)(9.8 m/s²)

<u>d = 0.87 m</u>

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(25 kg) (1 m) = 25 J

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From the calculations, the value of the acceleration due to gravity is 0.38 m/s^2.

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How many lenses with different focal lengths can be obtained by combining two surfaces whose radii of curvature are 4.00 cm and
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The lenses with different focal length are four.

Explanation:

Given that,

Radius of curvature R₁= 4

Radius of curvature R₂ = 8

We know ,

Refractive index of glass = 1.6

When, R₁= 4, R₂ = 8

We need to calculate the focal length of the lens

Using formula of focal length

\dfrac{1}{f}=(n-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}+\dfrac{1}{8})

\dfrac{1}{f}=\dfrac{9}{40}

f=4.44\ cm

When , R₁= -4, R₂ = 8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}+\dfrac{1}{8})

\dfrac{1}{f}=-\dfrac{3}{40}

f=-13.33\ cm

When , R₁= 4, R₂ = -8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}-\dfrac{1}{8})

\dfrac{1}{f}=\dfrac{3}{40}

f=13.33\ cm

When , R₁= -4, R₂ = -8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}-\dfrac{1}{8})

\dfrac{1}{f}=-\dfrac{9}{40}

f=-4.44\ cm

Hence, The lenses with different focal length are four.

8 0
3 years ago
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