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AnnZ [28]
3 years ago
8

Suppose you're performing experiments in science class in which you start with 70 bacteria and the amount of bacteria triples ev

ery hour. Write a function to represent the growth of the bacteria overtime in your science experiment
Physics
2 answers:
barxatty [35]3 years ago
5 0
1h----------------> 70x3=210 bacteria
2h-----------------> 210*3=630 bactaeria
let be y the number of bacteria at the t=0h
it is y=70 3^0
for t= 1h
y=70*3^1=210
for t=2h
y=70*3^2=630

so we can write y=70*3^x, where x is the number of hour


Brrunno [24]3 years ago
5 0

The function to represent the growth of bacteria will be

a=70*3^b

a is the number of bacteria at time b

b is the number of hour(time)

Explanation:

As per the given information You start your experiment with 70 bacteria,and the bacteria triples in every hour so in

1 hour = 70*3=210 bacteria are formed

2 hour = 210*3=630 bacteria are formed

3 hour=630*3=1,890 bacteria are formed

let a  be the number of bacteria at the t=1 hour

if  t= 1 hour then

a=70*3^1=210

if t=2 hour then

a=70*3^2=630

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Answer: C) The kinetic energy and the entropy of the water molecules increases through fusion.

Explanation:

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3 years ago
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two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i
NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

h = H - ut - \frac{1}{2}gt^2

0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 - 14.7t - 4.9t^2

4.9t^2 + 14.7t - 19.6 = 0

t^2 + 3t - 4 = 0

(t + 4)(t - 1) = 0

(t - 1) = 0

\boxed {t = 1 ~ second}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 + 14.7t - 4.9t^2

4.9t^2 - 14.7t - 19.6 = 0

t^2 - 3t - 4 = 0

(t - 4)(t + 1) = 0

(t - 4) = 0

\boxed {t = 4 ~ seconds}

The difference in the two ball's time in the air is:

\Delta t = 4 ~ seconds - 1 ~ second

\large {\boxed {\Delta t = 3 ~ seconds} }

<h2>Question B:</h2><h3>First Ball</h3>

v^2 = u^2 - 2gH

v^2 = (-14.7)^2 + 2(-9.8)(-19.6)

v^2 = 600.25

v = \sqrt {600.25}

\boxed {v = 24.5 ~ m/s}

<h3>Second Ball</h3>

v^2 = u^2 - 2gH

v^2 = (14.7)^2 + 2(-9.8)(-19.6)

v^2 = 600.25

v = \sqrt {600.25}

\boxed {v = 24.5 ~ m/s}

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

h = H - ut - \frac{1}{2}gt^2

h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2

\boxed {h = 11.025 ~ m}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2

\boxed {h = 25.725 ~ m}

The difference in the two ball's height after 0.500 s is:

\Delta h = 25.725 ~ m - 11.025 ~ m

\large {\boxed {\Delta h = 14.7 ~ m} }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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<u>We are given:</u>

Mass of Tarzan before swinging = 80 kg

Mass of Tarzan when swinging = 80 + 15 = 95 kg

Velocity of Tarzan = 7 m/s

The height of the rock Tarzan's monkey is sitting on = 3 m

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<u>Momentum of Tarzan before swinging:</u>

We know that:

Momentum = Mass*Velocity

Momentum = 80 * 7

Momentum = 560 kg m/s

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<u>Speed of Tarzan after grabbing the bananas:</u>

The momentum of Tarzan will remain the same but his mass will increase

So, Since Momentum = New Mass* velocity

560 = 95 * v                           [where v is the velocity of Tarzan]

v = 5.9 m/s

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<h3><u>Finding the Initial and Final KE and PE:</u></h3>

<em>Here, KE = Kinetic Energy and PE = Potential Energy</em>

<u>Initial and Final KE:</u>

We know that KE = 1/2*(mv²)

<u>Initial KE:</u>

Initial KE = 1/2*(mv²)          [where v is the velocity after picking the bananas]

Initial KE = 1/2*(95*5.9²)

Initial KE = 1653.5 Joules

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Final KE = 1/2*(mv²)            

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Since Tarzan will be at rest at the maximum point of the swing, v = 0 m/s

Final KE = 1/2*(95*0²)

Final KE = 0 Joules

<u>Initial and Final PE:</u>

We know that:

PE = mgh                

<em>[where g is the acceleration due to gravity and h is the height]</em>

<u>Initial PE:</u>

Since the height of Tarzan from the ground was 0 m at the beginning of the swing, h = 0

Initial PE = 95*10*0

Initial PE = 0 Joules

<u>Final PE:</u>

Let the maximum height of Tarzan be h m

Final PE = 95*10*h

Final PE = 950(h)

__________________________________________________________

<h3><u>Finding the maximum height Tarzan will reach:</u></h3>

<em>Here, KE = Kinetic Energy and PE = Potential Energy</em>

From the law of conservation of momentum, we know that:

Initial KE + Initial PE = Final KE + Final PE

Replacing the variables:

1653.5 + 0 = 0 + 950h

1653.5 = 950h

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