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2 years 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:

AVprozaik [17]2 years ago

7 0

The answer is true. Distraction “latency” lasts for about 27 seconds.

This means that even after driver put down the phone or stop fooling with the navigation system; he or she isn’t fully committed with the driving task. Talking on a cell phone and texting are frequent what people associate with distracted driving, but there are so much more activities behind distracted driving.

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Determine the scalar components Ra and Rb of the force R along the nonrectangular axes a and b. Also determine the orthogonal pr

Liula [17]

Answer: Hello your question is incomplete attached below is the complete question

Ra = 1132 N

Rb = 522.6 N

Pa = 679.7 N

Explanation:

To determine the scalar components Ra

$\frac{Ra}{Sin 120^o} = \frac{750}{sin 35^o}$

therefore : Ra = $\frac{sin120^o * 750}{sin 35^o}$ = 1132 N

To determine the scalar component Rb

$\frac{Rb}{sin 25^o} = \frac{750}{sin 35^o}$

therefore : Rb = $\frac{sin 25^o * 750}{sin 35^o}$ = 522.6 N

To determine the orthogonal projection Pa of R onto

Pa = 750 cos $25^o$ = 679.7 N

6 0

1 year ago

A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin

KengaRu [80]

a) 219.8 rad/s

b) $20.0 rad/s^2$

c) $2.9 m/s^2$

d) $7005 m/s^2$

e) Towards the axis of rotation

f) $0 m/s^2$

g) 31.9 m/s

Explanation:

a)

The angular velocity of an object in rotation is the rate of change of its angular position, so

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time elapsed

In this problem, we are told that the maximum angular velocity is

$\omega_{max}=35 rev/s$

The angle covered during 1 revolution is

$\theta=2\pi$ rad

Therefore, the maximum angular velocity is:

$\omega_{max}=35 \cdot 2\pi = 219.8 rad/s$

b)

The angular acceleration of an object in rotation is the rate of change of the angular velocity:

$\alpha = \frac{\Delta \omega}{t}$

where

$\Delta \omega$ is the change in angular velocity

t is the time elapsed

Here we have:

$\omega_0 = 0$ is the initial angular velocity

$\omega_{max}=219.8 rad/s$ is the final angular velocity

t = 11 s is the time elapsed

Therefore, the angular acceleration is:

$\alpha = \frac{219.8-0}{11}=20.0 rad/s^2$

c)

For an object in rotation, the acceleration has two components:

- A radial acceleration, called centripetal acceleration, towards the centre of the circle

- A tangential acceleration, tangential to the circle

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

Here we have

$\alpha =20.0 rad/s^2$

d = 29 cm is the diameter, so the radius is

r = d/2 = 14.5 cm = 0.145 m

So the tangential acceleration is

$a_t=(20.0)(0.145)=2.9 m/s^2$

d)

The magnitude of the radial (centripetal) acceleration is given by

$a_c = \omega^2 r$

where

$\omega$ is the angular velocity

r is the radius of the circle

Here we have:

$\omega_{max}=219.8 rad/s$ is the angular velocity when the fan is at full speed

r = 0.145 m is the distance of the gum from the centre of the circle

Therefore, the radial acceleration is

$a_c=(219.8)^2(0.145)=7005 m/s^2$

e)

The direction of the centripetal acceleration in a rotational motion is always towards the centre of the axis of rotation.

Therefore also in this case, the direction of the centripetal acceleration is towards the axis of rotation of the fan.

f)

The magnitude of the tangential acceleration of the fan at any moment is given by

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

When the fan is rotating at full speed, we have:

$\alpha=0$ , since the fan is no longer accelerating, because the angular velocity is no longer changing

r = 0.145 m

Therefore, the tangential acceleration when the fan is at full speed is

$a_t=(0)(0.145)=0 m/s^2$

g)

The linear speed of an object in rotational motion is related to the angular velocity by the formula:

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular velocity

r is the radius

When the fan is rotating at maximum angular velocity, we have:

$\omega=219.8 rad/s$

r = 0.145 m

Therefore, the linear speed of the gum as it is un-stucked from the fan will be:

$v=(219.8)(0.145)=31.9 m/s$

7 0

2 years ago

When a particle is a distance r from the origin, its potential energy function is given by the equation U(r)=kr, where k is a co

Reika [66]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.

5 0

2 years ago

A car of mass 1100kg moves at 24 m/s. What is the braking force needed to bring the car to a halt in 2.0 seconds? N

LenaWriter [7]

13200N

Explanation:

Given parameters:

Mass = 1100kg

Velocity = 24m/s

time = 2s

unknown:

Braking force = ?

Solution:

The braking force is the force needed to stop the car from moving.

Force = ma = $\frac{mv}{t}$

m is the mass of the car

v is the velocity

t is the time taken

Force = $\frac{1100 x 24}{2}$ = 13200N

Learn more:

Force brainly.com/question/4033012

#learnwithBrainly

8 0

1 year ago

5. A 1-kg car and a 2-kg car are both released from the top of the same hill and roll down a frictionless track. At the bottom o

grigory [225]

The cars will have equal speeds and the 2 kg car will have greater kinetic energy.

7 0

1 year ago