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2 years ago

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A 13.0 kg wheel, essentially a thin hoop with radius 1.80 m, is rotating at 469 rev/min. It must be brought to a stop in 16.0 s.

(a) How much work must be done to stop it? (b) What is the required average power? Give absolute values for both parts.

1 answer:

Stella [2.4K]2 years ago

7 0

Answer:

Explanation:

Given

mass of wheel m=13 kg

radius of wheel=1.8 m

N=469 rev/min

$\omega =\frac{2\pi \times 469}{60}=49.11 rad/s$

t=16 s

Angular deceleration in 16 s

$\omega =\omega _0+\alpha \cdot t$

$\alpha =\frac{\omega }{t}=\frac{49.11}{16}=3.069 rad/s^2$

Moment of Inertia $I=mr^2=13\times 1.8^2=42.12 kg-m^2$

Change in kinetic energy =Work done

Change in kinetic Energy $=\frac{I\omega ^2}{2}-\frac{I\omega _0^2}{2}$

$\Delta KE=\frac{42.12\times 49.11^2}{2}=50,792.34 J$

(a)Work done =50.79 kJ

(b)Average Power

$P_{avg}=\frac{E}{t}=\frac{50.792}{16}=3.174 kW$

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Svetradugi [14.3K]

Explanation:

The ratio of the areas is the square of the ratio of the radii.

A/A = 3.16² = 9.99

The ratio of the volumes is the cube of the ratio of the radii.

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3 years ago

A mug rests on an inclined surface, as shown in (Figure 1) , θ=17∘.

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Refer to the figure shown below.

g = 9.8 m/s², the acceleration due to gravity.
W = mg, the weight of the mug.
θ = 17°, the angle of the ramp.

Let μ = the coefficient of static friction.

The force acting down the ramp is
F = W sin θ = W sin(17°) = 0.2924W N
The normal reaction is
N = W cosθ = W cos(17°) = 0.9563W N
The resistive force due to friction is
R = μN = 0.9563μW N

For static equilibrium,
μN = F
0.9563μW =0.2924W
μ = 0.3058

The frictional force is F = μN = 0.2924W
The minimum value of μ required to prevent the mug from sliding satisfies
the condition
R > F
0.9563μW > 0.2924W
μ > 002924/.9563 = 0.306

Answer:
The frictional force is 0.2924mg, where m = the mass of the mug.
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2 years ago

Read 2 more answers

A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

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3 years ago

At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 44.0 N at the s

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Answer:

Weight at the surface of Jupiter's moon Io is 8.13 N .

Explanation:

Given :

Acceleration due to gravity at the surface of Jupiter's moon is $g_m=1.81\ m/s^2$ .

Weight of watermelon in earth , $W=44\ N$ .

Acceleration due to gravity at the surface of earth is $g=9.81\ m/s^2$ .

We know , weight is given by :

$W=mg\\m=\dfrac{W}{g}\\\\m=4.49\ kg$

Therefore , mass at the surface of Jupiter's moon Io is :

$W_m=mg_m\\\\W_m=4.49\times 1.81\\\\W_m=8.13 \ N$

Hence , this is the required solution .

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2 years ago