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LenaWriter [7]
2 years ago
11

A 9.0 kg potted plant slides down a 25.0º incline with an acceleration of 2.4 m/s2. What is the coefficient of kinetic friction

between the potted plant and the incline?
Physics
1 answer:
mezya [45]2 years ago
7 0

Answer:

0.196

Explanation:

According to Newton's second law

\sumFx = ma

Fm-Ff = ma

Fm is the moving force

Ff is the frictional force

m is the mass = 9kg

a is the acceleration = 2.4m/s²

Fm = Wsin theta

Fm = 9(9.8)sin25°

Fm = 37.28N

Ff is the frictional force = nmgcos theta

Ff = n(9)(9.8)cos25°

Ff = 79.94n

Substitute the given values into the formula

37.28-79.94n = 9(2.4)

-79.94n = 21.6-37.28

-79.94n = -15.68

n = 15.68/79.94

n = 0.196

Hence the coefficient of kinetic friction between the potted plant and the incline is 0.196

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Answer:

0.782 s

Explanation:

The water flows horizontally from the hose, so its initial vertical velocity is 0.

Given:

y₀ = 3 m

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v₀ = 0 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

0 m = 3 m + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 0.782 s

Round as needed.

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A kite 100 ft above the ground moves horizontally at a speed of 11 ft/s. At what rate is the angle (in radians) between the stri
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X^2+ 100^2=z^2

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Taking derivatives in terms of t

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Determine the binding energy per nucleon of an mg-24 nucleus. the mg-24 nucleus has a mass of 24.30506. a proton has a mass of 1
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The mass of Mg-24 is 24.30506 amu, it contains 12 protons and 12 neutrons.

Theoretical mass of Mg-24:

The theoretical mass of Mg-24 is:

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Neutron mass = 12 x 1.008665 amu = 12.104 amu

Theoretical mass = Hydrogen atom mass + Neutron mass = 24.1913 amu

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2 years ago
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