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Artyom0805 [142]

2 years ago

14

A 3.00 cm diameter coin rolls up a 30.0 degree incline plane. The coin starts with an initial angular speed of 60.0 rad/s in a s

traight line without slipping. How far does it roll up the incline?

1 answer:

yanalaym [24]2 years ago

8 0

Answer:

The coin roll up to 11.57 cm at 30° incline plane.

Explanation:

Given that,

Diameter = 3.00 cm

Angle = 30.0 °

Angular speed = 60.0 rad/s

We need to calculate the length

Using conservation of energy

$\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2=mgh$

$\dfrac{1}{2}mr^2\omega^2+\dfrac{1}{2}\times(0.4mr^2)\times\omega^2=mgl\sin\theta$

Put the value into the formula

$\dfrac{1}{2}\times(1.5\times10^{-2})^2\times(60.0)^2+\dfrac{1}{2}\times0.4\times(1.5\times10^{-2})^2\times(60.0)^2=9.8\times l\sin30$

$l\sin30=\dfrac{0.567}{9.8}$

$l=2\times\dfrac{0.567}{9.8}$

$l=0.1157\ m$

Hence, The coin roll up to 11.57 cm at 30° incline plane.

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The displacement of a 500 g mass, undergoing simple harmonic motion, is defined by the function :

Delicious77 [7]

The maximum kinetic energy, maximum potential energy and the maximum mechanical energy are equal to 7.56J.

<h3>What is simple harmonic motion?</h3>

Simple harmonic motion, in physics, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side.

Simple Harmonic Motion

The given equation of the simple harmonic motion is

$x=3.5 sin (\frac{\pi }{2t} + \frac{5\pi }{4} )$

Data;

ω = π/2

k = 1.254N/m

Solving this

$\frac{dx}{dt} = -3.5 X \frac{\pi }{2} cos (\frac{x\pi t}{2}+\frac{5\pi }{4} )$

Let's calculate the maximum velocity.

$V_{m} =\frac{3.5\pi }{2}$

This is only possible when cos θ = -1

The maximum kinetic energy is

$K_m =\frac{1}{2} mv^2 = \frac{1}{2} X \frac{500}{1000} X \frac{7^2\pi ^2}^{4} ^2$

$w^2 = \frac{k}{m} \\k = w^2m\\k = \frac{\pi ^2}{4} X \frac{500}{1000} \\k =1.254 N/m$

Using the value of spring constant, we can find the maximum potential energy.

$P.E =\frac{1}{2} k x^2\\P.E =\frac{1}{2} X 1.234 X 3.5^2 \\P.E = 7.56 J$

The maximum potential energy is 7.56J

The maximum mechanical energy is equal to the sum of maximum potential energy and the maximum kinetic energy.

ME = K.E + P.E

ME = 7.56J

From the calculations above, the maximum kinetic energy, maximum potential energy and the maximum mechanical energy are equal to 7.56J.

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8 0

1 year ago

A 295-kg object and a 595-kg object are separated by 4.10 m.

kodGreya [7K]

Answer:

a)F=3 x 10⁻⁷ N

b)x=2.405 m

Explanation:

Given that

m₁=295 kg

m₂=595 kg

d= 4.1 m

a)

m₃=63 kg

r=d/2 = 2.05 m

The force between the mass m₁ and m₃

$F_{13}=\dfrac{Gm_1m_3}{r^2}$

by putting the values

$F_{13}=\dfrac{Gm_1m_3}{r^2}$

$F_{13}=\dfrac{6.67\times 10^{-11}\times 295\times 63 }{2.05^2}$

F₁₃=2.94 x 10⁻⁷ N

The force between the mass m₂ and m₃

by putting the values

$F_{23}=\dfrac{Gm_2m_3}{r^2}$

$F_{23}=\dfrac{6.67\times 10^{-11}\times 595\times 63 }{2.05^2}$

F₂₃=5.94 x 10⁻⁷ N

The net force F

F= F₂₃- F₁₃

F=5.94 x 10⁻⁷ N-2.94 x 10⁻⁷ N

F=3 x 10⁻⁷ N

b)

Lest take at distance x from mass m₂ net force is zero.

$F_{23}=\dfrac{Gm_2m_3}{x^2}$

$F_{13}=\dfrac{Gm_1m_3}{(4.1-x)^2}$

Form above two equation

$\dfrac{Gm_1m_3}{(4.1-x)^2}=\dfrac{Gm_2m_3}{x^2}$

$\dfrac{m_1}{(4.1-x)^2}=\dfrac{m_2}{x^2}$

$\dfrac{295}{(4.1-x)^2}=\dfrac{595}{x^2}$

x²=2.01(4.1-x)²

x=1.42 (4.1-x)

x=5.82 - 1.42x

x=2.405 m

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2 years ago

Two satellites of equal mass are orbiting the planet Mars. Satellite B is closer to Mars than Satellite A. How does the gravitat

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The object in the lower orbit will move faster in its orbit, and will have a shorter orbital period. None of that depends on their masses either.

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2 years ago

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mote1985 [20]

Answer:

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here we know

speed = 343 m/s

wavelength = 0.77955 m

now frequency is given by

$f = \frac{343m/s}{ 0.77955m}$

f = 440 Hz

2. Frequency = speed / wavelength

here we know

speed = 343 m/s

wavelength = 0.52028 m

now frequency is given by

$f = \frac{343m/s}{ 0.52028m}$

f = 659.3 Hz

3. Frequency = speed / wavelength

here we know

speed = 343 m/s

wavelength = 0.65552 m

now frequency is given by

$f = \frac{343m/s}{ 0.65552m}$

f = 523.25 Hz

4. Wavelength = speed / Frequency

here we know

speed = 343 m/s

Frequency = 587.33 Hz

now wavelength is given by

$\lambda = \frac{343m/s}{587.33}$

$\lambda = 0.584 m$

5. Wavelength = speed / Frequency

here we know

speed = 343 m/s

Frequency = 493.88 Hz

now wavelength is given by

$\lambda = \frac{343m/s}{493.88}$

$\lambda = 0.6945 m$

6. Wavelength = speed / Frequency

here we know

speed = 343 m/s

Frequency = 698.46 Hz

now wavelength is given by

$\lambda = \frac{343m/s}{698.46}$

$\lambda = 0.491 m$

7. Frequency = speed / wavelength

here we know

speed = 343 m/s

wavelength = 0.5840 m

now frequency is given by

$f = \frac{343m/s}{0.5840m}$

f = 587.3 Hz

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here we know

speed = 343 m/s

wavelength = 0.4375 m

now frequency is given by

$f = \frac{343m/s}{0.4375m}$

f = 784 Hz

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here we know

speed = 343 m/s

Frequency = 783.99 Hz

now wavelength is given by

$\lambda = \frac{343m/s}{783.99}$

$\lambda = 0.4375 m$

10. Wavelength = speed / Frequency

here we know

speed = 343 m/s

Frequency = 659.26 Hz

now wavelength is given by

$\lambda = \frac{343m/s}{659.26}$

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4 0

2 years ago

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