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2 years ago

Consider the position vs. time graph below for a woman's movement in a hallway. What is the woman's velocity from 4 to 5 s?

Physics

1 answer:

Ksenya-84 [330]2 years ago

7 0

Answer:

The answer is " $6\ \frac{m}{s}$ "

Explanation:

The formula for velocity:

$\to \overline{v}={\frac{\Delta x}{\Delta t}}$

$=\frac{6}{1}\\\\=6\ \frac{m}{s}$

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Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0 °C for the following reacti

Yakvenalex [24]

The equilibrium constant of the reaction at 25⁰c will be 426827.5.

Theory-

Equilibrium constant :The equilibrium constant comes from the chemical equilibrium law. For the chemical equilibrium state, at a fixed constant temperature, the ratio of the product of the reaction's multiplication to the concentration of its reactants' multiplication, and each is raised to the power to the corresponding coefficients of the elements in the reaction.

The chemical equilibrium is given by for a general chemical reaction.

a. A+ b. B ⇌ c. C+ d. D,.

Kc =[C]c [D]d/[A]a [B]b.

Gibb's free energy :The second law of thermodynamics can be arranged in such a way that it gives a new expression when a chemical reaction happens at a constant temperature and constant pressure.

G=H-TS

Calculations:-

T=25⁰c

G=51.4 x 10³J

$\\\\k=GR+\frac{nRT}{Z} \\$

k= equilibrium constant ,G=Gibbs free energy ,n= no. of moles ,R=Gas constant ,T=temperature ,Z=compressibility

$Ideal.Situation=\left \{ {{Z=1} \atop {n=1}} \right.$

$\\\\\\k=GR+RT$

k=51.4 x 10³ x 8.3 + 8.3 x 25

k=426827.5

To learn equilibrium constant-

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6 0

1 year ago

Train cars are coupled together by being bumped into one another. Suppose two loaded cars are moving toward one another, the fir

tia_tia [17]

Answer:

7560 Joules

Explanation:

$m_1$ = Mass of first car = $1.5\times 10^5\ kg$

$m_2$ = Mass of second car = $2\times 10^5\ kg$

$u_1$ = Initial Velocity of first car = 0.3 m/s

$u_2$ = Initial Velocity of second car = -0.12 m/s

v = Velocity of combined mass

As linear momentum of the system is conserved

$m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{1.5\times 10^5\times 0.3 + 2\times 10^5\times -0.12}{1.5\times 10^5 + 2\times 10^5}\\\Rightarrow v=0.06\ m/s$

Energy lost is

$\Delta E=\Delta E_i-\Delta E_f\\\Rightarrow \Delta=\frac{1}{2}(m_1u_1^2 + m_2u_2^2-(m_1+m_2)v^2)\\\Rightarrow \Delta=\frac{1}{2}(1.5\times 10^5\times 0.3^2 + 2\times 10^5\times (-0.12)^2-(1.5\times 10^5 + 2\times 10^5)\times 0.06^2)\\\Rightarrow \Delta=7560\ J$

The Energy lost in the collision is 7560 Joules

7 0

3 years ago

Suppose you want to design an air bag system that can protect the driver at a speed 100 km/h (60 mph) if the car hits a brick wa

34kurt

When solving question that contains equations and the use mathematical computations, It is always ideal to list the parameters given.

Now, given that:

the speed of the car which is the initial velocity (u) = 100 km/h before it hits the wall.
after hitting the wall, the final velocity will be (v) = 0 km/h

Assumptions:

Suppose we make an assumption that the distance travelled during the collision of the car with the brick wall (S) = 1 m
That the car's acceleration is also constant.

∴

For a motion under constant acceleration, we can apply the kinematic equation:

$\mathsf{v^2 = u^2 + 2as}$

where;

v = final velocity

u = initial velocity

a = acceleration

s = distance

From the above equation, making acceleration (a) the subject of the formula:

$\mathsf{v^2 - u^2 =2as }$

$\mathsf{a = \dfrac{v^2 - u^2 }{2s}}$

The initial velocity (u) is given in km/h, and we need to convert it to m/s as it has an effect on the unit of the acceleration.

since 1 km/h = 0.2778 m/s

100 km/h = 27.78 m/s

$\mathsf{a = \dfrac{(0)^2 - (27.78)^2 }{2(1)}}$

$\mathsf{a = \dfrac{- 771.7284 }{2}}$

a = - 385.86 m/s²

Similarly, from the kinematic equation of motion, the formula showing the relation between time, acceleration and velocity is;

v = u + at

where;

v = 0

-u = at

$\mathsf{t = \dfrac{-u}{a}}$

$\mathsf{t = \dfrac{-27.78}{-385.86}}$

t = 0.07 seconds

An airbag is designed in such a way as to prevent the driver from hitting on the steering wheel or other hard substance that could damage the part of the body. The use of the seat belt is to keep the driver in shape and in a balanced position against the expansion that occurred by the airbag during the collision on the brick wall.

Thus, we can conclude that the airbag must be inflated at 0.07 seconds faster before the collision to effectively protect the driver.

Learn more about the kinematic equation here:

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3 0

3 years ago

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).

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

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3 years ago

Many galaxies appear to have supermassive black holes in their centers powering active galactic nuclei (also called AGN). The Sc

Lena [83]

Answer:i dont remember

Explanation: ?>>> confused

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3 years ago