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18 days ago

What is the angular acceleration of the pencil when it makes an angle of 10.0 degrees with the vertical?

Physics

2 answers:

Keith_Richards [2.2K]18 days ago

8 0

When the pencil is positioned at an angle of 10.0 degrees to the vertical, its angular acceleration is 1306.6 rad / s^2

Clarification:

Imagine balancing a pencil on its tip. If you let it fall, what is the angular acceleration when it is at a 10.0-degree angle from the vertical? Express your answer in radians per second squared.

When the pencil is in balance, there is no net torque; however, once released, the weight of the pencil creates a net torque.

We fundamentally understand the connection between torque and angular acceleration. Thus, we find that the angular acceleration of the pencil is established.

The pencil's length L is 15 cm and its weight (m) corresponds to the angle (θ) of 10° from vertical. Hence, the torque (τ) caused by its weight (W) is articulated as:

$\tau = r xF\\\tau =\frac{L}{2} W sin \theta\\\tau =\frac{L}{2} mg sin 10^o$

$\tau =\frac{0.15}{2} m*10* sin 10^o\\\tau = 0.075 m*10* sin 10^o\\\tau = 0.130m$

From there, we can calculate the angular acceleration:

$\tau = I \alpha = F r\\\tau = 0.130m = F r\\\alpha = \frac{0.098}{0.000075} = 1306.6 rad / s^2$

To delve deeper into angular acceleration,

Sav [2.2K]18 days ago

6 0

<span> </span><span>When the net torque and moment of inertia are given, calculating becomes straightforward.

Using the equation torque = I * alpha, where I represents the moment of inertia and alpha is the angular acceleration.

Consequently, 0.098 / 0.000075 results in 1306.666... rad / s^2

While the angular acceleration stays the same, you can also determine the angular velocity at that instance, which is 21.36 rad / s.</span>

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An axle passes through a pulley. Each end of the axle has a string that is tied to a support. A third string is looped many time

Keith_Richards [2263]

Answer:

ΔL = MmRgt / (2m + M)

Explanation:

The system starts from rest, so the change in angular momentum correlates directly to its final angular momentum.

ΔL = L − L₀

ΔL = Iω − 0

ΔL = ½ MR²ω

To determine the angular velocity ω, begin by drawing a free body diagram for both the pulley and the block.

For the block, two forces act: the weight force mg downward and tension force T upward.

For the pulley, three forces are present: weight force Mg down, a reaction force up, and tension force T downward.

For the sum of forces in the -y direction on the block:

∑F = ma

mg − T = ma

T = mg − ma

For the sum of torques on the pulley:

∑τ = Iα

TR = (½ MR²) (a/R)

T = ½ Ma

Substituting gives:

mg − ma = ½ Ma

2mg − 2ma = Ma

2mg = (2m + M) a

a = 2mg / (2m + M)

The angular acceleration of the pulley is:

αR = 2mg / (2m + M)

α = 2mg / (R (2m + M))

Finally, the angular velocity after time t is:

ω = αt + ω₀

ω = 2mg / (R (2m + M)) t + 0

ω = 2mgt / (R (2m + M))

Substituting into the previous equations gives:

ΔL = ½ MR² × 2mgt / (R (2m + M))

ΔL = MmRgt / (2m + M)

3 0

1 month ago

the flow energy of 124 L/min of a fluid passing a boundary to a system is 108.5 kJ/min. Determine the pressure at this point

serg [2593]

Answer:

The pressure measured at this moment is 0.875 mPa

Explanation:

Given that,

Flow energy = 124 L/min

Boundary to system P= 108.5 kJ/min

$P=1.81\ kW$

We are tasked with finding the pressure here

Applying the pressure formula

$P=F\times v$

$P=A_{1}P_{1}\times v_{1}$

Here, $A_{1}v_{1}=Q_{1}$

Where, v refers to velocity

Insert the values into the equation

$1.81 =P_{1}\times0.124\times\dfrac{1}{60}$

$P_{1}=\dfrac{1.81\times60}{0.124}$

$P_{1}=875.80\ kPa$

$P_{1}=0.875\ mPa$

Therefore, the pressure at this moment is 0.875 mPa

5 0

1 month ago

A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sph

Softa [2035]

B. The charge on A is -q; B has no charge. Given that a positive charge is situated at the center of an uncharged metallic sphere which is insulated and disconnected from the ground, a negative charge (-q) will appear on the inner surface A of the sphere. Should the exterior surface B be grounded, it will become neutral, resulting in no charge remaining on surface B.

4 0

3 days ago

A 0.050-kg lump of clay moving horizontally at 12 m/s strikes and sticks to a stationary 0.15-kg cart that can move on a frictio

Ostrovityanka [2208]

Answer:

Explanation:

According to the parameters provided,

mass of the clay lump, m₁ = 0.05 kg

initial velocity of the lump, u₁ = 12 m/s

mass of the cart, m₂ = 0.15 kg

initial speed of the cart, u₂ = 0

As the clay adheres to the cart, we have an inelastic collision scenario. Let v represent the combined speed of both the cart and lump post-collision. Given that momentum is conserved, we have:

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$v=\dfrac{0.05\ kg\times 12\ m/s+0}{0.05\ kg+0.15\ kg}$

The resultant speed is v = 3 m/s.

Thus, the final speed of both cart and lump following the collision is 3 m/s. This concludes the solution.

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19 days ago

Below you are given data about a wave in three different substances.

inna [2210]

1) The wave's period remains constant across different media

2) The wave's velocity varies depending on the medium it travels through

3) As a wave transitions between media, its speed, direction, and wavelength can change, while its frequency stays unchanged

Clarification:

The period of a wave signifies the duration it takes for one full oscillation.

The wave's period is the inverse of its frequency:

$T=\frac{1}{f}$

where

T denotes the period

f is the frequency

The provided table illustrates that the frequency remains consistent across the three media; hence, the period is unchanged as it solely relies on frequency. We can compute it as we know that

f = 350 Hz

thus the period equals

$T=\frac{1}{350}=2.86\cdot 10^{-3} s = 2.86 ms$

The velocity of a wave can be derived from the wave equation:

$v=f \lambda$

where

f indicates the frequency

$\lambda$ is the wavelength

$f=350 Hz, \lambda = 0.75 m$ , resulting in a speed of

$v_1 = (350)(0.75)=262.5 m/s$

In the second medium,

$f=350 Hz, \lambda = 0.70 m$ , leading to a speed of

$v_2 = (350)(0.70)=245 m/s$

In the third medium,

$f=350 Hz, \lambda = 0.65 m$ , showing a speed of

$v_3 = (350)(0.65)=227.5 m/s$

As a result, we conclude that the wave's speed varies with the medium.

- The wave's direction alters. Specifically, if the subsequent medium is of greater optical density, the wave bends towards the normal; conversely, it bends away if the second medium is of lesser optical density.

- The wave's speed is affected. The wave decelerates in media with higher optical density and accelerates in those with lower optical density.

- The wave's frequency remains unchanged.

- Ultimately, the wave's wavelength is modified. If moving into a medium of greater optical density, the wavelength decreases, while it increases in one of lower optical density.

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