Find the acceleration of a train whose speed increases from 9 m/s to 22 m/s in 160s

Answers

Answer 1
Answer: acceleration = (v2-v1) / (t2-t1) = (22-9) / 160 = 13/160 m/s

Related Questions

What is the value of current when the voltage is 8 volts
What is cosmic background radiation?
A 10 kg bowling ball would require what force to accelerate down an alleyway at a rate of 3 m/s2? F= M= A=
Which of the following changes would make a heat engine less efficient
What is the apparent magnitude of the brightest star in the Big Dipper?

What is the ultimate source of heat and light for the solar system? Option a) The Moon Option b) The Earth Option c) The Sun Option d) The Milky Way

Answers

Answer:C) THE SUN

Explanation:SUN IS THE PRIMARY SOURCE OF ENERGY IN OUR SOLAR SYSTEM

A motorist traveling at 17 m/s encounters a deer in the road 39 m ahead. If the maximum acceleration the vehicle’s brakes are capable of is −7 m/s 2 , what is the maximum reaction time of the motorist that will allow her or him to avoid hitting the deer? Answer in units of s. 019 (part 2 of 2) 10.0 points If his or her reaction time is 1.21983 s, how fast will (s) he be traveling when (s)he reaches the deer? Answer in units of m/s.

Answers

Answer:

1.     t_reaction = 1.08 s

2.     v₀₁ = 16.365 m/s

Explanation:

1. This is a kinematics exercise, let's analyze the situation a bit, we can calculate the braking distance and the rest of the distance we can use to calculate the reaction time.

Braking distance

           v² = v₀² + 2 a x

when he finishes braking the speed is v = 0

            0 = v₀² + 2 a x

            x = -v₀² / 2a

            x = - 17²/2 (-7)

            x = 20.64 m

the distance for the reaction is

            d = x_reaction + x

            x_reaction = d - x

            x_reaction = 39 - 20.64

            x_reaction = 18.36 m

as long as it has not reacted the vehicle speed is constant

            v = x_reaction / t_reaction

            t_reaction = x_reaction / v

            t_reaction = 18.36 / 17

            t_reaction = 1.08 s

2. Let's find the distance traveled in the reaction time of t1 = 1.21983 s

       as the speed is constant

           v = x / t

           x₁ = v t₁

the distance traveled during braking is

           v² = v₀² + 2a x₂

           0 = v₀² + 2 a x₂

           x₂ = -v₀² / 2a

         

           v = v₀

the total distance is

         x_total = x₁ + x₂

         x_total = v₀ t₁ + v₀² / 2a

         39 = v₀ 1.21983 + v₀²/14

         v₀² + 17.08 vo - 546 =0

we solve the second degree equation

       v₀ = [ -17.08 ±√(17.08²  + 4  546) ]/2

       v₀ = [-17.08 ± 49.81 ]/2

       v₀₁ = 16.365 m/s

       v₀₂ = - 33.445 m/s

as the acceleration is negative the correct result is v₀₁ = 16.365 m/s

In a titration, the point where is indicator changes color and stays that way is the?

Answers

It is the end point of a reaction

C12H22O11 + 11H2SO4 12C + 11H2SO4 + 11H2OHow do you know the concentrated sulfuric acid is a catalyst?

Answers

By looking at the equation 
C12H22O11 + 11H2SO4 12C + 11H2SO4 + 11H2O
we can conclude that the concentrated sulfuric acid is a catalyst because
the sulfuric acid doesn't react with the reactant and it is not consumed.

Hope this helps! If you have any other questions or would like further explanation just let me know!! :)

Answer:no

Explanation:

A cow standing atop a building in Times Square recalled a funny joke and began to laugh. The uncontrollable laughter caused the cow to fall over the side of the building. He fell for a time period of 3.5 s and landed in a bed of bushes. How fast was the cow moo-ving when he reached the bushes?Which equation should be used to solve the problem?

Answers

Answer:

Vf = 34.3 m/s

1st equation of motion was used to solve.

Explanation:

In order to find the final speed of the cow, when it hits the bushes, we can use first equation of motion:

Vf = Vi + gt

where,

Vf = Final Velocity of Cow = ?

Vi = Initial Velocity of Cow = 0 m/s

g = acceleration due to gravity = 9.8 m/s²

t = time taken = 3.5 s

Therefore,

Vf = 0 m/s + (9.8 m/s²)(3.5 s)

Vf = 34.3 m/s

1st equation of motion was used to solve.

What will happen to the pitch of a sound as that sound’s source approaches an observer? Explain why this happens, based on what you have learned about wave properties. Please use 3 content related sentences

Answers

The pitch of a sound is decreased, when a sound source approaches the observer.

Explanation:

Pitch is nothing but it is a property that helps to determine whether the sound is high or low. Doppler effect helps to understand this concept easily. The pitch of the sound is mainly calculated based on the frequency of the sound wave arose from the sound source.

As the wellspring of sound waves moves toward an audience, the sound waves draw nearer together, expanding their recurrence and the pitch of the sound.