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Scorpion4ik [409]
3 years ago
15

The stoplights on a street are designed to keep traffic moving at 24 mi/h. The average length of a street block between traffic

lights is about 80 m. What must be the time delay between green lights on successive blocks to keep the traffic moving continuously? There are 1.609 × 103 m in a mile. Answer in units of s.
Physics
1 answer:
Crazy boy [7]3 years ago
5 0

Answer:

t = 7.5 s

Explanation:

As we know that the average speed between two lights is given as

v = 24 mph

here we know that

v = 24 \times \frac{1602 m}{3600 s}

v = 10.68 m/s

now the distance between two lights is 80 m

so here if all the cars are moving continuously between two lights then the time delay between two lights is given as

t = \frac{L}{v}

t = \frac{80}{10.68}

t = 7.5 s

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1. Vpa = 180m/s. @ 0 deg.
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<span> Vpg = Vpa + Vag,
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3.  1 hr = 3600s. <span>d = Vt = 163.7m/s * 3600s = 589,320m.

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A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall, and the other end is connec
Wewaii [24]

Answer:

Velocity = 0.4762 m/s

Explanation:

Given the details for the simple harmonic motion from the question as:

Angular frequency, ω = 12 rad/s

Amplitude, A = 0.060 m

Displacement, y = 0.045 m

The initial Energy =  U  = (1/2) kA²    

where A is the amplitude and k is the spring constant.

The final energy is potential and kinetic energy

   K + U =   (1/2) mv²   + (1/2) kx²  

where  x  is the displacement

m is the mass of the object

v is the speed of the object

Since energy is conservative. So, the final and initial energies are equal  as:

   (1/2) k A²   = (1/2) m v²   + (1/2) kx²  

Using,   ω² = k/m, we get:  

Velocity:

v=\omega\times \sqrt{[ A^2 - y^2 ]}

v=\omega\times \sqrt{[ {0.06}^2 - {0.045}^2 ]}

<u>Velocity = 0.4762 m/s</u>

4 0
3 years ago
You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s
marta [7]

1) Acceleration of the car in front: -7.89 m/s^2

The only data we need for this part of the problem is:

u = 27.7 m/s --> initial velocity of the car

\mu=0.804 --> coefficient of friction between the car wheels and the road

From the coefficient of friction, we can find the deceleration of the car. In fact, the force of friction is given by

F=-\mu mg

where m is the car's mass and g=9.81 m/s^2 is the acceleration due to gravity. We can find the car's acceleration by using Newton's second law:

a=\frac{F}{m}=\frac{-\mu mg}{m}=\mu g=(0.804)(9.81 m/s^2)=-7.89 m/s^2

And the negative sign means it is a deceleration.


2) Braking distance for the car in front: 48.6 m

This can be found by using the following SUVAT equation:

v^2 - u^2 = 2aS

where

v=0 is the final velocity of the car

u=27.7 m/s is the initial velocity of the car

a=-7.89 m/s^2 is the acceleration of the car

S is the braking distance

By re-arranging the formula, we find S:

S=\frac{v^2-u^2}{2a}=\frac{0-(27.7 m/s)^2}{2(-7.89 m/s^2)}=48.6 m


3) Minimum safe distance at which you can follow the car: 15.0 m

In this case, we must calculate the thinking distance, which is the distance you travel before hitting the brakes. During this time, the speed of your car is constant, so the thinking distance is given by

d_t = ut=(27.7 m/s)(0.543 s)=15.0 m

After hitting the brakes, your car decelerates at the same rate of the car in front of you, so the braking distance is the same of the other car:

d_b=48.6 m

So the total distance your car covers is

S'=d_t+d_b=15.0 m +48.6 m=63.6 m

At the same time, the car in front of you just covered a distance of 48.6 m. So, in order to avoid the collision, you should travel at a distance equal to

d=S'-S=63.6 m-48.6 m=15.0 m


6 0
3 years ago
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