PHYSICS COLLEGE

Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting line 1.00 s after Stan does. Stan moves with a constant acceleration of 3.1 m/s2 while Kathy maintains an acceleration of 4.99 m/s. 2 (a) Find the time at which Kathy overtakes Stan. s from the time Kathy started driving (b) Find the distance she travels before she catches him (c) Find the speeds of both cars at the instant she overtakes him. Kathy m/s Stan m/s

Answers

Answer 1

Answer:

Answer:

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

(b) d=37.36 m

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 = 19.31m/s : Kathy's final speed

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Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

The answer is

$T_2 = 1.008$ % higher than $T_1$

$T_2 = 0.99$ % lower than $T_1$

Explanation:

From the question we are told that

The first string has a frequency of $f_1 = 230 Hz$

The period of the beat is $t_(beat) = 0.99s$

Generally the frequency of the beat is

$f_(beat) = (1)/(t_(beat))$

Substituting values

$f_(beat) = (1)/(0.99)$

$= 1.01 Hz$

From the question

$f_2 - f_1 = f_(beat)$ for $f_2$ having a higher tension

$f_2 - 230 = 1.01$

$f_2 = 231.01Hz$

From the question

$(f_2)/(f_1) = \sqrt{(T_2)/(T_1) }$

$(T_2)/(T_1) = (f_2^2)/(f_1^2)$

Substituting values

$(T_2)/(T_1) = ((231.01)^2)/((230)^2)$

$T_2 = 1.008$ % higher than $T_1$

For $f_2$ having a lower tension

$f_1 - f_2 = f_(beat)$

$230 - f_2 = 1.01$

$f_2 = 230 -1.01$

$= 228.99$

From the question

$(f_2)/(f_1) = \sqrt{(T_2)/(T_1) }$

$(T_2)/(T_1) = (f_2^2)/(f_1^2)$

Substituting values

$(T_2)/(T_1) = ((228.99)^2)/((230)^2)$

$T_2 = 0.99$ % lower than $T_1$

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Answers

Answer:

The answer is below

Explanation:

The speed of the boat in still water is perpendicular to the speed of the water flow. Therefore the speed relative to the ground (V), the speed of flow and the speed of the boat in still water form a right angled triangle. Hence the speed relative to the ground is given as:

V² = 56² + 126²

V² = 19012

V = 137.9 m/s

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Answers

Answer:

The electric flux remains unchanged

Explanation:

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C. 97 dB
D. 223 dB
E. 179 dB

Answers

Answer:

Explanation:

We have given the intensity of the loud car horn $I=0.005w/m^2$

We know that $I_O=10^(-12)w/m^2$

Now the sound intensity level is given by $\beta =10log(I)/(I_0)=10log(0.005)/(10^(-12))=96.98dB$ , which is nearly equal to 97

So the sound intensity level will be 97 dB

So option (c) will be the correct option

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Answers

Answer:

0.83x10^-9 T

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Explanation:

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Therefore,

B = E/c = 0.25 ÷ 3x10^8

B = 0.83x10^-9 T

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