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3 years ago

5

A carousel - a horizontal rotating platform - of radius r is initially at rest, and then begins to accelerate constantly until i

t has reached angular velocity after 2 complete revolutions. What is the angular acceleration of the carousel during this time

1 answer:

OLga [1]3 years ago

7 0

Answer:

α = (ω²)/8π

Explanation:

The angular acceleration(α) of the carousel can be determined by using rotational kinematics:

ω² =ωo² + 2αθ

Let's make α the subject of this equation ;

ω² - ωo² = 2αθ

α = (ω² −ωo²)/2θ

Now, from the question, since initially at rest, thus, ωo = 0

Also,since 2 revolutions, thus, θ = 2 x 2π = 4π since one revolution is 2π

Plugging in the relevant values to get ;

α = (ω²)/2(4π)

α = (ω²)/8π

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Arigid body must rotate about an axis in order for it to have angular momentum about that axis. True False

kompoz [17]

Answer:

False

Explanation:

Let's consider the definition of the angular momentum,

$\vec{L} = I \vec{\omega}$

where $I = \int\limits_m r^2 dm = \lim_{n \to \infty} \sum\limits_{i=1}^n m_i r_i^2$ is the moment of inertia for a rigid body. Now, this moment of inertia could change if we change the axis of rotation, because "r" is defined as the distance between the puntual mass and the nearest point on the axis of rotation, but still it's going to have some value. On the other hand,

$\vec{\omega} = \frac{\vec{r} \times \vec{v}}{r^2}$ so $\vec{\omega} \neq 0$ unless $\vec{r}$ ║ $\vec{v}$ .

In conclusion, a rigid body could rotate about certain axis, generating an angular momentum, but if you choose another axis, there could be some parts of the rigid body rotating around the new axis, especially if there is a projection of the old axis in the new one.

7 0

2 years ago

What distance does electromagnetic radiation travel in 55.0 μs ?

Komok [63]

Electromagnetic radiation is an energy that is known as light. so electromagnetic radiation will have the same speed as the speed of light which is 3 x 10^8 m/s. so the distance it travel at 55 x 10^-6 s is:
D = ( 3 x 10^8 m/s ) ( 55 x 10^-6 s )
D = 16500 m

4 0

2 years ago

A uniform log of length L is inclined 30° from the horizontal when supported by a frictionless rock located 0.6L from its left e

mafiozo [28]

Answer:

x = 0.974 L

Explanation:

given,

length of inclination of log = 30°

mass of log = 200 Kg

rock is located at = 0.6 L

L is the length of the log

mass of engineer = 53.5 Kg

let x be the distance from left at which log is horizontal.

For log to be horizontal system should be in equilibrium

∑ M = 0

mass of the log will be concentrated at the center

distance of rock from CM of log = 0.1 L

now,

∑ M = 0

$m_{log} g \times 0.1 L = m_{engineer} g \times (x - 0.6 L)$

$200 \times 0.1 L = 53.5 \times (x - 0.6 L)$

$0.374 L =x - 0.6 L$

x = 0.974 L

hence, distance of the engineer from the left side is equal to x = 0.974 L

7 0

3 years ago

Light of wavelength 476.1 nm falls on two slits spaced 0.29 mm apart. What is the required distance from the slits to the screen

stich3 [128]

Answer:

The distance is $D = 2.6 \ m$

Explanation:

From the question we are told that

The wavelength of the light is $\lambda = 476.1 \ nm = 476.1 *10^{-9} \ m$

The distance between the slit is $d = 0.29 \ mm = 0.29 *10^{-3} \ m$

The between the first and second dark fringes is $y = 4.2 \ mm = 4.2 *10^{-3} \ m$

Generally fringe width is mathematically represented as

$y = \frac{\lambda * D }{d}$

Where D is the distance of the slit to the screen

Hence

$D = \frac{y * d}{\lambda }$

substituting values

$D = \frac{ 4.2 *10^{-3} * 0.29 *10^{-3}}{ 476.1 *10^{-9} }$

$D = 2.6 \ m$

7 0

3 years ago

A rifle is aimed horizontally at a target 49 m away. the bullet hits the target 2.3 cm below the aim point. part a what was the

Nezavi [6.7K]

We use the equation of motion for vertical component,

$s_{y} = u_{y} t+\frac{1}{2} gt^2.$

Here, $s_{y}$ is displacement of bullet, $u_{y}$ is vertical initial velocity of bullet which is equal to zero because bullet was fired horizontally, and t is time of flight.

Therefore,

$s_{y} =\frac{1}{2}gt^2$

Given, $s_{y} =2.3 \ cm = 2.3 \times 10^{-2} \ m$

Substituting the values, we get time of flight

$2.3 \times 10^{-2} \ m = \frac{1}{2} \times 9.8 \ m/s^2 \times t^2 \\\\ t =\sqrt{46.94 \times 10^{-4} \ s } = 0.069 \ s$

4 0

3 years ago