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Zina [86]
3 years ago
6

A rocket is fired from rest from the ground (y = 0) at time t0 = 0 s. As the rocket is burning its fuel, it moves vertically upw

ard with a constant acceleration of 3.25 m/s2 . At t1 = 10.0 s, all the fuel has been used up and the rocket is in free fall. Air resistance can be neglected up until t3 = 17.5 s when the rocket deploys a parachute. At this time the rocket immediately reaches its terminal velocity; this means that the rocket is no longer accelerating.
1a. (10 points) Find the maximum height of the rocket with respect to the ground. How long after being launched does it take for the rocket to reach this height?
1b. (10 points) How long after being launched does it take for the rocket to return to the ground? What is the rocket’s displacement over this entire trip? What is the rocket’s average velocity over this entire trip? What is the rocket’s average speed over this entire trip? What is the rocket’s average acceleration over this entire trip? Be sure to include both magnitude and direction for any vector quantities.
1c. (10 points) What is the rocket’s displacement from ta = 12.0 s to tb = 20.0 secs? What is the rocket’s average velocity from ta to tb? What is the rocket’s average speed from ta to tb? What is the rocket’s average acceleration from ta to tb? Be sure to include both magnitude and direction for any vector quantities.
1d. (10 points) Sketch a y v. t plot for the motion of the rocket. Be sure to label the positions of the rocket at t0 = 0 s, t1 = 10.0 s, t2 (when the rocket is at its highest point), t3 = 17.5 s, and t4 (when the rocket returns to the ground).

Physics
1 answer:
elixir [45]3 years ago
6 0

Answer:

Explanation:

attached is the solution

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the ground is BREAKING

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Motion maps for two objects, Y & Z, are shown.
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Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting li
Fofino [41]

Answer:

(a) t=3.87 s :time at which Kathy overtakes Stan

(b) d=37.36 m

(c) vf₁ = 15.097 m/s : Stan's final speed

    vf₂ = 19.31 m/s : Kathy's final speed

Explanation:

kinematic analysis

Because Kathy and Stan move with uniformly accelerated movement we apply the following formulas:

vf= v₀+at Formula (1)

vf²=v₀²+2*a*d Formula (2)

d= v₀t+ (1/2)*a*t² Formula (3)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Nomenclature

d₁: Stan displacement   

t₁ :  Stan time

v₀₁: Stan initial speed

vf₁: Stan final speed

a₁:  Stan acceleration

d₂: car displacement   

t₂ : Kathy time

v₀₂: Kathy initial speed

vf₂: Kathy final speed

a₂:  Kathy acceleration

Data

v₀₁ = 0

v₀₂ = 0

a₁ = 3.1 m/s²

a₂= 4.99 m/s²

t₁ = (t₂ +1) s

Problem development

By the time Kathy overtakes Stan, the two will have traveled the same distance:

d₁ = d₂

t₁ = (t₂ +1)

We aplpy the Formula (3)

d₁ = v₀₁t₁ + (1/2)*a₁*t₁²

d₁ = 0 + (1/2)*(3.1)*t₁²

d₁ =  1.55*t₁² ; Stan's cinematic equation 1

d₂ = v₀₂t₂ + (1/2)*a₂*t₂²

d₂ = 0 + (1/2)*(4.99)*t₂²

d₂ = 2.495* t₂² : Kathy's cinematic equation 2

d₁ = d₂

equation 1 = equation 2

1.55*t₁²  =  2.495* t₂²  , We replace t₁ = (t₂ +1)

1.55* (t₂ +1) ² = 2.495* t₂²

1.55* (t₂² +2t₂+1) = 2.495* t₂²

1.55*t₂²+1.55*2t₂+1.55 = 2.495* t₂²

1.55t₂²+3.1t₂+1.55=2.495t₂²

(2.495-1.55)t₂² - 3.1t₂ - 1.55 = 0

0.905t₂² - 3.1t₂ - 1.55 = 0  Quadratic equation

Solving the quadratic equation we have:

(a) t₂ = 3.87 s : time at which Kathy overtakes Stan

(b) Distance in which Kathy catches Stan

we replace t₂ = 3.87 s in equation 2

d₂ = 2.495*( 3.87)²

d₂ = 37.36 m

(c) Speeds of both cars at the instant  Kathy overtakes Stan

We apply the Formula (1)

vf₁= v₀₁+a₁t₁    t₁ =( t₂+1 ) s=( 3.87 + 1 ) s = 4.87 s

vf₁= 0+3.1* 4.87

vf₁ = 15.097 m/s : Stan's final speed

vf₂ = v₀₂+a₂ t₂  

vf₂ =0+4.99* 3.87

vf₂ = 19.31 m/s : Kathy's final speed

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What is the definition of a neutron?
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Answer:

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(a) Amplitude=1.5 m

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Wavelength is the length of one complete wave or distance traveled by a wave in a time equal to Time Period. Here wavelength= 2 m

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f=7.5 Hz

so frequency=7.5 Hz

(d) Time period= 0.133 s

Time period= T= 1/f

T= 1/7.5

T=0.133 s

so time period=0.133 s

5 0
3 years ago
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