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den301095 [7]
3 years ago
6

An object with a mass of 32 kg has an initial energy of 500 J. At the end of the experiment, the velocity of the object is recor

ded as 5.1 m/s. If the object travelled 50 m to get to this point, what was the average force of friction on object during the trip? Assume no potential energy. Show all work
Physics
1 answer:
Nezavi [6.7K]3 years ago
3 0

Explanation:
the concept of conservation of the mechanical nerve
initial
Em₀ = 500 J
The energy is totally kinetic
Em₀ = K = ½ m v₀²
v₀ =
v₀ = √ (2 500/32)
v₀ = 5.59 m / s
v² = v₀² - 2 a x
the negative sign is because its stopping
a =
a = (5.59² - 5.1²) / 2 50
a = 0.0524 m / s²
Newton's second law
F = ma
F = 32 0.0524
F = 1.68 N
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Answer:

% reduction in area = 54.26 %

percentage elongation = 43.16 %

Explanation:

a) percentage reduction in area = \frac{(A_1 - A_2)}{A_1 } * 100

A_1  =\pi r^2 = \pi * 5.88^2=108.8 mm2

A_2=\pi * 3.98^2= 49.97 mm2

% reduction in area =\frac{(108.8 - 49.97)}{108.8} * 100 = 54.26 %

b)percentage elongation = \frac{(66 - 46.1)}{46.1} *100 = 43.16 %

5 0
3 years ago
You stand on a frictional platform that is rotating at 1.8 rev/s. Your arms are outstretched, and you hold a heavy weight in eac
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Answer:

20.62361 rad/s

489.81804 J

Explanation:

I_i = Initial moment of inertia = 9.3 kgm²

I_f = Final moment of inertia = 5.1 kgm²

\omega_i = Initial angular speed = 1.8 rev/s

\omega_f = Final angular speed

As the angular momentum of the system is conserved

I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s

The resulting angular speed of the platform is 20.62361 rad/s

Change in kinetic energy is given by

\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J

The change in kinetic energy of the system is 489.81804 J

As the work was done to move the weight in there was an increase in kinetic energy

6 0
3 years ago
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QveST [7]

Answer:

270 m/s²

Explanation:

Given:

α = 150 rad/s²

ω = 12.0 rad/s

r = 1.30 m

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a

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The tangential component is:

at = αr

at = (150 rad/s²)(1.30 m)

at = 195 m/s²

The radial component is:

ar = v² / r

ar = ω² r

ar = (12.0 rad/s)² (1.30 m)

ar = 187.2 m/s²

So the magnitude of the total acceleration is:

a² = at² + ar²

a² = (195 m/s²)² + (187.2 m/s²)²

a = 270 m/s²

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