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1 month ago

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Question 5 of 5: Someone texting or talking spans an average of 27 seconds after they put the phone down are still thinking abou

t what they just did is called latency.

1 answer:

Yuliya22 [3.3K]1 month ago

7 0

The statement is correct. The period of distraction, referred to as "latency," persists for roughly 27 seconds.

This indicates that after a driver has set down their phone or ceased interacting with the navigation system, they remain partially disengaged from the driving task. While use of cell phones and texting are commonly linked to distracted driving, there are numerous other behaviors that contribute to this issue.

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Consider a perfectly insulated cup (no

ValentinkaMS [3465]

Answer:

When ice is subjected to heat, it melts; however, the temperature remains constant at 0◦ C.

Explanation:

Solution

The heat supplied by the heater is solely utilized for the melting of the ice, thus maintaining the temperature at 0◦ C.

Once all the ice has liquefied, the temperature of the resulting water will start to rise over time.

Note: please see the attached document with solutions featuring diagrams related to this explanation

7 0

2 months ago

In a novel from 1866 the author describes a spaceship that is blasted out of a cannon with a speed of about 11.000 m/s. The spac

Sav [3153]

Answer:

$a=0.284\ m/s^2$

Explanation:

We start with the fact that

Initially, the spacecraft was at rest, u = 0

The final velocity of the rocket is given as v = 11 m/s

The distance that the rocket covers during acceleration is given as d = 213 m

We seek to determine the acceleration that the occupants of the spacecraft experience during launch, which is derived from the principles of kinematics. By applying the

third motion equation we can find the acceleration.

$v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(11)^2-(0)^2}{2\times 213}\\\\a=0.284\ m/s^2$

Thus, the acceleration felt by those inside the spacecraft is $0.284\ m/s^2$ .

5 0

28 days ago

Derive an algebraic equation for the vertical force that the bench exerts on the book at the lowest point of the circular path i

Keith_Richards [3271]

a)

i) 120 s

ii) 1.57 m/s

b)

i) Refer to the attached diagram

ii) Up

c) $N=mg+m\frac{v_b^2}{R}$

d) Greater than

Explanation:

The problem does not provide full details: consult the attachments for the complete text.

a)

The revolution period of the book equals the total duration needed for the book to make one full revolution.

By examining the graph, we can approximate the revolution period by calculating the time difference between two successive points of the book's motion that share the same shape.

We could use the time difference between two adjacent crests to estimate the period. The first crest is observed at t = 90 s, and the following crest appears at t = 210 s.

This results in the revolution period being

T = 210 - 90 = 120 s

ii)

The tangential speed of the book is computed as the ratio of the distance traveled over one revolution (i.e., the circumference of the wheel) to the revolution period.

Mathematically:

$v_b=\frac{2\pi R}{T}$

where

R represents the wheel radius

T = 120 s indicates the period

Based on the graph, the book reaches a maximum at x = +30 m and a minimum at x = -30 m, giving the diameter of the wheel as

d = +30 - (-30) = 60 m

This means the radius calculates to

R = d/2 = 30 m

So, the final speed is

$v_b=\frac{2\pi (30)}{120}=1.57 m/s$

b)

i) Please consult the attached free-body diagram for the book when at its lowest point.

Two forces act on the book at the lowest position:

- The weight of the book, represented as

$W=mg$

where m denotes the book's mass and g stands for gravitational acceleration. This force functions downward.

- The normal force the bench exerts on the book is represented by N. This force acts upward.

ii)

While at its lowest position, the book maintains a horizontal motion at constant speed.

Nevertheless, the book is undergoing acceleration. Acceleration is defined as the rate of velocity change, which is vectorial, having both speed and direction. While the speed remains unchanged, the direction changes (upward), indicating the book has upward net acceleration.

According to Newton's second law, the net vertical force acting on the book corresponds with the vertical acceleration:

$F=ma$

where F = net force, m = mass, a = acceleration. Thus, if a is non-zero, the upward net force must exist in line with the direction of the acceleration.

c)

As discussed in part b), there are two forces influencing the book at the lowest point:

- The weight, $W=mg$ , directed downward

- The normal force from the bench, N, directed upward

Given that the book is in uniform circular motion, the net force must match the centripetal force $m\frac{v_b^2}{R}$ , leading us to the equation:

$N-mg=m\frac{v_b^2}{R}$

where

$v_b$ represents the speed of the book

R stands for the radius of the circular path.

We derive an expression for the normal force:

$N=mg+m\frac{v_b^2}{R}$

d)

As per the discussions in parts c) and d):

- The normal force acting on the book at its lowest point becomes

$N=mg+m\frac{v_b^2}{R}$

- The weight (gravitational force) of the book is

$W=mg$

Upon comparing these two equations, we conclude:

$N>W$

Thus, it is evident that the normal force exerted by the bench exceeds the weight of the book.

4 0

2 months ago

Keisha told her friend that she can get to the town library by walking one mile east, turning left, and then walking a half a mi

Maru [3345]

To enhance these instructions effectively and without difficulty, Keisha should specify the initial point from which the directions commence. Is the journey beginning at Keisha's residence? Perhaps it starts at her friend's home? Or is it from my place? Understanding the reference frame for coordinates is crucial and significantly impacts comprehension.

7 0

1 month ago

Read 2 more answers

A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceler

Maru [3345]

Answer:

Part a)

$t_1 = \frac{\omega_1}{\alpha}$

Part b)

$\theta = \frac{1}{2}\alpha t_1^2$

Part c)

$t = \frac{t_1}{5}$

Explanation:

Part a)

Since there is a consistent torque, the time required to achieve acceleration is expressed as

$\omega_f = \omega_i + \alpha t$

$\omega_1 = 0 + \alpha t_1$

$t_1 = \frac{\omega_1}{\alpha}$

Part b)

Angular displacement can be determined as

$\theta = \omega_i t_1 + \frac{1}{2}(\alpha) t_1^2$

$\theta = 0 + \frac{1}{2}\alpha t_1^2$

$\theta = \frac{1}{2}\alpha t_1^2$

Part c)

The brakes create a certain angular deceleration that is calculated as

$\alpha_d = - 5\alpha$

Now we find

$\omega_f = \omega _i + \alpha t$

$0 = \omega_1 - 5 \alpha_1 t$

$t = \frac{\omega_1}{5 \alpha}$

Moreover, we know that

$t_1 = \frac{\omega_1}{\alpha}$

therefore, we derive

$t = \frac{t_1}{5}$

6 0

2 months ago