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3 years ago

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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

ne 1.00 s after Stan does. Stan moves with a constant acceleration of 3.4 m/s2 while Kathy maintains an acceleration of 4.49 m/s2. (a) Find the time at which Kathy overtakes Stan. 6.7 Correct: Your answer is correct. seenKey 6.7 s from the time Kathy started driving

1 answer:

Alekssandra [29.7K]3 years ago

3 0

Answer:

time at which Kathy overtakes Stan is 6.70 sec

Explanation:

given data

time = 1 sec

acceleration= 3.4 m/s²

acceleration = 4.49 m/s²

to find out

the time at which Kathy overtakes Stan

solution

we consider here travel time for kathy = t1

and travel time for stan is = t2

and we know initial velocity = 0

so

t1 = 1 + t2

and distance travel equation by kinematic is

d1 = ut + $\frac{1}{2}$ at²

d1 = 0+ $\frac{1}{2}$ 4.49 (t2)² ................1

and

d2 = 0 + $\frac{1}{2}$ 3.4 (1+t2)² ..................2

and when overtake distance same so from equation 1 and 2

$\frac{1}{2}$ 4.49 (t2)² = $\frac{1}{2}$ 3.4 (1+t2)²

t2 = 6.703828 sec

so time at which Kathy overtakes Stan is 6.70 sec

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Answer:

a) $\Delta s=443037.9747\ J.K^{-1}$

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Explanation:

a)

Given:

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final temperature of car, $T_f=44.4+273=317.4\ K$

We know that the change in entropy is given by:

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)}$ (heat is transferred into the system of car)

$\Delta s=443037.9747\ J.K^{-1}$

b)

amount of heat transfer form the system of house, $dQ=5.8\times 10^8\ J$

initial temperature of house, $T_i=23.5+273=296.5\ K$

final temperature of house, $T_f=5.3+273=278.3\ K$

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{5.8\times 10^8}{278.3-296.5}$

$\Delta s=31868131.8681\ J.K^{-1}$

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3 years ago

How much work is done on a 75 newton bowling ball when you carry it horizontally across a 10 meter room

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F = force applied to hold the weight of the bowling ball = weight of the bowling ball = 75 N

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θ = angle between the force in vertically upward direction and displacement in horizontal direction = 90

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3 years ago

Question 9 of 27

Ksju [112]

The power of man performing 500 J of work in 8 seconds is 62.5 J/s.

Power can be defined as the pace at which work is completed in a given amount of time.

Horsepower is sometimes used to describe the power of motor vehicles and other machinery.

The pace at which work is done on an item is defined as its power. Power is a temporal quantity.

Which is connected to how quickly a project is completed.

The power formula is shown below.

Power = Energy / Time

Power = E / T

Because the standard metric unit for labour is the Joule and the standard metric unit for time is the second, the standard metric unit for power is a Joule / second, defined as a Watt and abbreviated W.

Here we have given Energy as 500 J and Time as 8 second.

Power = Energy / Time

Power = 500 / 8 Joule / sec

Power = 250 / 4 Joule / sec

Power = 125 / 2 Joule / sec

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Power came out to be 62.5 J/s when the man performed 500 Joule of work in 8 seconds.

So we can conclude that the power in the Energy transmitted per unit of time, and can be find out by dividing Energy by time. In our case the Power came out to be 62.5 Joule / Second.

Learn more about Power here:

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2 years ago

An observer stands on the side of the front of a stationary train. When the train starts moving with constant acceleration, the

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To solve this problem we will use the linear motion description kinematic equations. We will proceed to analyze the general case by which the analysis is taken for the second car and the tenth. So we have to:

$x = v_0 t \frac{1}{2} at^2$

Where,

x= Displacement

$v_0$ = Initial velocity

a = Acceleration

t = time

Since there is no initial velocity, the same equation can be transformed in terms of length and time as:

$L = \frac{1}{2} a t_1 ^2$

For the second cart

$2L \frac{1}{2} at_2^2$

When the tenth car is aligned the length will be 9 times the initial therefore:

$9L = \frac{1}{2} at_3^2$

When the tenth car has passed the length will be 10 times the initial therefore:

$10L = \frac{1}{2}at_4^2$

The difference in time taken from the second car to pass it is 5 seconds, therefore:

$t_2-t_1 = 5s$

From the first equation replacing it in the second one we will have that the relationship of the two times is equivalent to:

$\frac{1}{2} = (\frac{t_1}{t_2})^2$

$t_1 = \frac{t_2}{\sqrt{2}}$

From the relationship when the car has passed and the time difference we will have to:

$(t_2-\frac{t_2}{\sqrt{2}}) = 5$

$t_2 (\sqrt{2}-1) = 3\sqrt{2}$

$t_2= (\frac{5\sqrt{2}}{\sqrt{2}-1})^2$

Replacing the value found in the equation given for the second car equation we have to:

$\frac{L}{a} = \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2$

Finally we will have the time when the cars are aligned is

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The time when you have passed it would be:

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The difference between the two times would be:

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Therefore the correct answer is C.

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3 years ago