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3 years ago

What is the final speed of a 60 kg boulder dropped from a 111 meter cliff

1 answer:

3 0

After rolling off the edge of the cliff and falling ' M ' meters down,
the speed of the boulder is

Square root of ( 19.6 M ) .

If M=111 meters, then the speed is <em>46.64 meters per second</em>.

We have known for roughly 500 years that if there's no air resistance,
the mass of the falling object makes no difference, and all objects fall
with the same acceleration, speed, time to splat, etc.

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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

With what tension must a rope with length 3.00 mm and mass 0.105 kgkg be stretched for transverse waves of frequency 40.0 HzHz t

VladimirAG [237]

Answer:

the tension of the rope is 34.95 N

Explanation:

Given;

length of the rope, L = 3 m

mass of the rope, m = 0.105 kg

frequency of the wave, f = 40 Hz

wavelength of the wave, λ = 0.79 m

Let the tension of the rope = T

The speed of the wave is given as;

$v = f\lambda = \sqrt{\frac{T}{\mu} } \\\\where;\\\\\mu \ is \ mass \ per \ unit \ length\\\\\mu = \frac{0.105}{3} = 0.035 \ kg/m\\\\v = f\lambda = 40 \times 0.79 = 31.6 \ m/s\\\\v = \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (31.6^2)(0.035)\\\\T = 34.95 \ N$

Therefore, the tension of the rope is 34.95 N

4 0

3 years ago

A big wheel at a theme park has a diameter of 14m and people on the ride complete one revolution in 24s. calculate the distance

just olya [345]

Explanation:

We'll call the radius r and the diameter d:

We also assume that the riders are at a distance r = d/2 = 7m from the center of the wheel.

The period of the wheel is 24s. The tangent velocity of the wheel (and the riders) will be: (2pi/T)*r = 0.8 m/s (circa).

It means that in 3 minutes (180 seconds) they'll run 0.8 m/s * 180s = 144m.

Hopefully I understood the question. If yes, that's the answer.

7 0

2 years ago

_____ are placed on dangerous machinery to detect motion, light, heat, pressure, or another stimulus. Their presence helps prote

goblinko [34]

Answer;

-Sensors

-Sensors are placed on dangerous machinery to detect motion, light, heat, pressure, or another stimulus. Their presence helps protect operators from injury while working on machines.

Explanation;

-Machinery, safety and factory floor sensors and switches help workers become more productive, efficient, and safe.

-Hazardous machines and systems are frequently equipped with safety elements (safety doors) with a locking mechanism to protect the operator. Their function is to prevent hazardous machine functions if the safety door is not closed and locked and to keep the safety door closed and locked until the risk of injury has passed.

4 0

3 years ago

Read 2 more answers

A patient has an ongoing history of cancer. She has a tumor in the abdominal region, and has been undergoing treatment for it. T

Vadim26 [7]

Answer:

the answer is at the BOTTOM OF THEIR QUESTION

Explanation:

IT IS CORRECT BTW

6 0

3 years ago