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Gre4nikov [31]
3 years ago
10

An inattentive driver is traveling 18.0 m/s when he notices a red light ahead. his car is capable of decelerating at a rate of 3

.35 m/s2 . if it takes him 0.200 s to get the brakes on and he is 20.0 m from the intersection when he sees the light, will he be able to stop in time?
Physics
1 answer:
ryzh [129]3 years ago
3 0

Speed of the car given initially

v = 18 m/s

deceleration of the car after applying brakes will be

a = 3.35 m/s^2

Reaction time of the driver = 0.200 s

Now when he see the red light distance covered by the till he start pressing the brakes

d_1 = v* t

d_1 = 18* 0.200 = 3.6 m

Now after applying brakes the distance covered by the car before it stops is given by kinematics equation

v_f^2 - v_i^2 = 2 a s

here

vi = 18 m/s

vf = 0

a = - 3.35

so now we will have

0^2 - 18^2 = 2*(-3.35)(s)

s = 48.35 m

So total distance after which car will stop is

d = d_1 + d_2

d = 48.35 + 3.6 = 51.95 m

So car will not stop before the intersection as it is at distance 20 m

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Answer:

Sound is produced when an object vibrates, creating a pressure wave.

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The winning time for a 500.0 mile circular race track was 3 hours and
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Explanation:

We have,

Distance traveled in a circular track is 500 miles

The winning time was 3 hours and  13 minutes. It means time is 3.217 hours.

The driver's average speed is given by total distance divided by total time taken. Its formula can be written as :

v=\dfrac{d}{t}\\\\v=\dfrac{500\ miles}{3.217\ h}\\\\v=155.42\ mph

At the end of the race, the driver reaches the point form where he has started. It means the displacement of the driver is equal to 0. Hence, driver's average velocity is equal to 0.

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3 years ago
The overhang beam is subjected to the uniform distributed load having an intensity of w = 50 kn/m. determine the maximum shear s
alex41 [277]
Given:

Uniform distributed load with an intensity of W = 50 kN / m on an overhang beam.

We need to determine the maximum shear stress developed in the beam:

τ = F/A

Assuming the area of the beam is 100 m^2 with a length of 10 m.

τ = F/A
τ = W/l
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τ = 5kN/m^2
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8 0
3 years ago
A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the lens strength (a.k.a, lens p
Kipish [7]

Answer:

20.0 cm

Explanation:

Here is the complete question

The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?

Solution

Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.

Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.

Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m

Now, P' = 1/u + 1/v

1/u = P'- 1/v

1/u = 55.0 D - 1/0.02 m

1/u = 55.0 m⁻¹ - 1/0.02 m

1/u = 55.0 m⁻¹ - 50.0 m⁻¹

1/u = 5.0 m⁻¹

u = 1/5.0 m⁻¹

u = 0.2 m

u = 20 cm

So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.

8 0
2 years ago
Mario rdoll a coin up a slope at 2 m/s. It travels 2.7 m, comes to a stop and rolls back down. What is the coin's acceleration?
leva [86]

The coin's acceleration is <u>0.37 m/s²</u>

Acceleration is the rate of change of the velocity of an item with appreciation to time. Accelerations are vector portions. The orientation of an item's acceleration is given by the orientation of the net pressure appearing on that object.

<u>Calculation:-</u>

<u />

V² = U -2aS

a = U/2S

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   =  <u>0.37 m/s²</u>

Acceleration is the charge at which velocity modifications with time, in terms of each speed and route. A factor or an object moving in a straight line is accelerated if it quickens or slows down. movement on a circle is extended despite the fact that the rate is consistent because the course is continually changing.

Learn more about acceleration here:- brainly.com/question/29110429

#SPJ9

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