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3 years ago

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How can a passenger tell when the seatbelt is properly used? when it clicks. when it is loosely draped across the lap. when it c

rossed the sternum and is snug across the lap. when the seatbelt is very tight across the sternum and lap?

1 answer:

babymother [125]3 years ago

6 0

When it crossed the sternum and is snug around the lap

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What is the angular displacement of the wheel between t = 5 s and t = 15 s?

Mkey [24]

The question seems incomplete. The complete text is:

a)What is the angular displacement of the wheel between t = 5 s and t = 15 s

b)What is the angular velocity of the wheel at 15 s

And it refers to the attached figure.

a) 25 rad

The graph shown represent the angular position of the wheel at different times.

Therefore, we can simply calculate the angular displacement between two times by calculating the difference between the angular position at t2 and the angular position at t1.

At $t_1 = 5 s$ , the angular position from the graph is $\theta_1 = 100 rad$

At $t_2 = 15 s$ , the angular position from the graph is $\theta_2 = 125 rad$

Therefore, the angular displacement is

$\Delta \theta= \theta_2 - \theta_1 = 125-100 = 25 rad$

2) -5.0 rad/s

For a angular displacement vs time graph, the angular velocity at any time is simply equal to the slope of the curve at that time.

Here we want to calculate the angular velocity at t = 15 s, so we have to calculate the slope at that time.

By noting that the slope is constant in the last part of the motion, we find that the slope between 10 s and 20 s is:

$\frac{\Delta \theta}{\Delta t}=\frac{100 rad - 150 rad}{20 s - 10 s}=-5.0 rad/s$

This slope is constant between 10 s and 20 s, so the angular velocity of the wheel at t = 15 s

$\omega = -5.0 rad/s$

5 0

3 years ago

Can a body possess velocity at the same time in horizontal and vertical directions?

iogann1982 [59]

Answer:

Yes

Explanation:

A body can possess velocity at the same time in horizontal and vertical direction

For example

A projectile

5 0

2 years ago

Derive the equation of motion of the block of mass m1 in terms of its displacement x. The friction between the block and the sur

Alenkasestr [34]

Answer:

the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$

the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be $\phi$ and $\beta$ respectively.

The velocity of the block to be = v

The equivalent mass of the system = $m_e$

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

$K.E = \frac{1}{2} m_ev^2$ --------------- equation (1)

The angular velocity of the cylinder = $\omega$ : &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

$v = \omega R$

$\omega =\frac{v}{R}$

The kinetic energy of the system in terms of individual mass can be written as:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2$

By replacing $\omega$ with $\frac{v}{R}$ ; we have:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2$

$K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2$ ------------------ equation (2)

Equating both equation (1) and (2); we have:

$m_e = m_1+m_2+\frac{I}{R^2}$

Therefore, the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$ which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

$f_{block }=m_1gsin \beta$

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

$f_{cylinder} = m_2gsin \phi$

Equating the above both equations; we have the equation of motion of the equivalent system to be

$m_e \bar x = f_{cylinder}-f_{block}$

which can be written as follows from the previous derivations

$(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Finally; the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

8 0

3 years ago

Which of the following is formed by constructive erosion? 1. delta 2. gully 3. rill 4. glacier

lana [24]

1. Delta, is formed by constructive erosion.

7 0

3 years ago

A professor sits at rest on a stool that can rotate without friction. The rotational inertia of the professor-stool system is 4.

Anestetic [448]

Answer:

$\omega=0.37 [rad/s]$

Explanation:

We can use the conservation of the angular momentum.

$L=mvR$

$I\omega=mvR$

Now the Inertia is I(professor_stool) plus mR², that is the momentum inertia of a hoop about central axis.

So we will have:

$(I_{proffesor - stool}+mR^{2})\omega=mvR$

Now, we just need to solve it for ω.

$\omega=\frac{mvR}{I_{proffesor-stool}+mR^{2}}$

$\omega=\frac{1.5*2.7*0.4}{4.1+1.5*0.4^{2}}$

$\omega=0.37 [rad/s]$

I hope it helps you!

5 0

3 years ago