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3 years ago

The of a wave is the density of the medium’s particles at the compression of the wave.

Physics

2 answers:

Westkost [7]3 years ago

7 0

'Amplitude' is the correct answer i just took the quiz

andrew11 [14]3 years ago

6 0

<span>Amplitude is the correct answer. I hope this helps.</span>

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Two identical carts are free to move along a straight frictionless track. At time t1, cart X is moving at 2.0 m/s when it collid

jarptica [38.1K]

Answer:

Explanation:

Using the law of conservation of momentum;

$m_1 u_1+m_2u_2 = m_1v_1+m_2v_2$

here;

There is a need for conservation of the total momentum that occurred before and after the collision.

So;

$m_1$ = mass of cart X

$m_2$ = mas 9f cart Y

$u_1$ = velocity of cart X (before collision)

$u_2$ = velocity of cart Y (before collision)

$v_1$ = velocity of cart X (after collision)

$v_2$ = velocity of cart Y (after collision)

So;

$m(u_1+0) =(m_1v+m_2)v$

because the mass is identical and v represents the velocity of both carts.

Now;

$u_1$ = 2 m/s

$u_2$ = 0 ( at rest)

∴

m(2) = (2m)v

v = 1 m/s

Thus, we can see from the graphical image attached below that the velocity of X reduces to 1 m/s after collision with cart Y.

8 0

3 years ago

Cardiovascuar exercise involves

bogdanovich [222]

Cardiovascular exercise involves movement that gets your heart rate up to improve oxygen consumption in the body.

Examples of cardiovascular exercise (Aerobic) include:
Spinning
Running
Swimming
Walking
Hiking
Dancing
Kick Boxing

4 0

3 years ago

A spring on a horizontal surface can be stretched and held 0.2 m from its equilibrium position with a force of 16 N. a. How much

nekit [7.7K]

Answer:

a $W_{3.5} = 490 \ J$

b $W_{2.5} = 250 \ J$

Explanation:

Generally the force constant is mathematically represented as

$k = \frac{F}{x}$

substituting values given in the question

=> $k = \frac{16}{0.2}$

=> $k = 80 \ N /m$

Generally the workdone in stretching the spring 3.5 m is mathematically represented as

$W_{3.5} = \frac{1}{2} * k * (3.5)^2$

=> $W_{3.5} = \frac{1}{2} * 80 * (3.5)^2$

=> $W_{3.5} = 490 \ J$

Generally the workdone in compressing the spring 2.5 m is mathematically represented as

$W_{2.5} = \frac{1}{2} * k * (2.5)^2$

=> $W_{2.5} = \frac{1}{2} * 80 * (2.5)^2$

=> $W_{2.5} = 250 \ J$

5 0

2 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).

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

7 0

3 years ago

Which of the following is NOT true of electromagnetic waves? A. An electric field is created in any region of space in which a m

Iteru [2.4K]

The incorrect statement about electromagnetic waves is C. induction of electric fields by changing magnetic fields only occurs if a conducting material is present.
Electromagnetic waves do not rely on any medium for propagation, which means that the generation of fields is irrespective of the presence of a conducting material.

6 0

3 years ago

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