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4 years ago

How much work is done by a car engine if the 1200kg car goes 13m up a hill and accelerates from 6 m/s to 7.5m/s?

1 answer:

7 0

The work done by the engine is equal to the gravitational potential energy (GPE) put into it. GPE = mgh
GPE = Work = 1200·9.81·13 = 153,036J

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Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v

mihalych1998 [28]

Answer:

$\displaystyle a(0 )=8.133\ m/s^2$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=0.52\ m/s^2$

b. $\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. $t=9.9 \ sec$

Explanation:

Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

$\displaystyle V(t)=a(1-e^{bt})$

For Carl Lewis's run at the 1987 World Championships, the values of a and b are

$\displaystyle a=11.81\ ,\ b=-0.6887$

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

$\displaystyle V(t)=11.81(1-e^{-0.6887t})$

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

$\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})$

$\displaystyle a(t)=8.133547\ e^{-0.6887t}$

For t=0

$\displaystyle a(0)=8.133547\ e^o$

$\displaystyle a(0 )=8.133\ m/s^2$

For t=2

$\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}$

$\displaystyle a(4)=0.52\ m/s^2$

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

$\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C$

$\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C$

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

$\displaystyle x(0)=0=>11.81\times1.45201+C=0$

Solving for C

$\displaystyle c=-17.1482\approx -17.15$

Now we complete the equation for the distance

$\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. Find the time Lewis needed to sprint 100.0 m.

The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

$\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100$

Rearranging

$\displaystyle t+1.45\ e^{-0.6887t}=9.92$

We define an auxiliary function f(t) to help us find the value of t.

$\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92$

Let's try for t=9 sec

$\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92$

Now with t=9.9 sec

$\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184$

That was a real close guess. One more to be sure for t=10 sec

$\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081$

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).

$At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}$

7 0

3 years ago

Where is potential energy in food stored

AlexFokin [52]

The potential energy is stored in the chemical bonds of the food. When those bonds break up during the metabolic processes, the energy is released. After that, that energy is stored in the Adenosine Triphosphate bonds aka ATP. The simplest way to think is to think of food as the tightly bound atoms. When the chemical bonds between those atoms break, the stored energy in that food is released.

6 0

3 years ago

If you have 80g of a radioactive substance whose half life is 2 days, how long will it take for the substance to decay to the po

Ber [7]

Answer:

6 days.

Explanation:

From radioactivity, The expression for half life is given as,

R/R' = 2⁽ᵃ/ᵇ)................... Equation 1

Where R = original mass of the radioactive substance, R' = Remaining mass of the radioactive substance after decay, a = Total time taken to decay, b = half life.

Given: R = 80 g, R' = 10 g, b = 2 days.

Substitute into equation 1

80/10 = 2⁽ᵃ/²⁾

8 = 2⁽ᵃ/²⁾

2³ = 2⁽ᵃ/²)

Equating the base and solving for a

3 = a/2

a = 2×3

a = 6 days.

5 0

3 years ago

Tara is an electrician. Which field of science does Tara need to know the most about in order to do her job?

katovenus [111]

Chemistry and physics

3 0

3 years ago

What is the half life-life of your 100 atoms of Carbon-14?

Naddika [18.5K]

Answer:

Answer: It takes 5,730 years for half the carbon-14 to change to nitrogen; this is the half-life of carbon-14. After another 5,730 years only one-quarter of the original carbon-14 will remain

3 0

3 years ago