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4 years ago

Which statement describes an unanticipated impact of a technological

1 answer:

6 0

Answer: the development of antibiotics allowed people to be treated for bacterial infections, but their overuse had led to strains of bacteria that cannot be killed by available antibiotics

Explanation: I just took the test

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A mass weighing 11 lb stretches a spring 4in. The mass is pulled down an additional 3 in and is then set in motion with an initi

Shtirlitz [24]

Answer:

a) $x(t) = 2.845\cdot \cos \left(1.732\cdot t -0.473\pi \right)$ , b) $T = 3.628\,s$ , $A = 2.845\,ft$ , $\phi = -0.473\pi$

Explanation:

a) The system mass-spring is well described by the following equation of equilibrium:

$\Sigma F = k\cdot x - m\cdot g = m\cdot a$

After some handling in physical and mathematical definition, the following non-homogeneous second-order linear differential equation:

$\frac{d^{2}x}{dt^{2}}+\frac{k}{m}\cdot x = g$

The solution of this equation is:

$x (t) = A\cdot \cos \left(\sqrt{\frac{k}{m} } \cdot t + \phi\right)$

The velocity function is:

$v(t) = \sqrt{\frac{k}{m} }\cdot A\cdot \sin \left(\sqrt{\frac{k}{m} }\cdot t +\phi \right)$

Initial conditions are:

$x(0\,s) = 0.25\,ft, v(0\,s) = -5\,\frac{ft}{s}$

Equations at $t = 0\,s$ are:

$0.25\,ft = A\cdot \cos \phi\\-5\,\frac{ft}{s} =\sqrt{\frac{k}{m} }\cdot A\cdot \sin \phi$

The spring constant is:

$k = \frac{11\,lbf}{0.333\,ft}$

$k = 33\,\frac{lbf}{ft}$

After some algebraic handling, amplitude and phase angle are found:

$\phi = -0.473\pi$

$A = 2.845\,ft$

The position can be described by this function:

$x(t) = 2.845\cdot \cos \left(1.732\cdot t -0.473\pi \right)$

b) The period of the motion is:

$T = \frac{2\pi}{\sqrt{\frac{k}{m} } }$

$T = 3.628\,s$

The amplitude is:

$A = 2.845\,ft$

The phase of the motion is:

$\phi = -0.473\pi$

8 0

4 years ago

In an 8.00km race, one runner runs at a steady 11.8 km/hr and another runs at 15.0 km/hr. How far from the finish line is the sl

Aleks [24]

Answer:

The slower runner is 1.71 km from the finish line when the fastest runner finishes the race.

Explanation:

Given;

the speed of the slower runner, u₁ = 11.8 km/hr

the speed of the fastest runner, u₂ = 15 km/hr

distance, d = 8 km

The time when the fastest runner finishes the race is given by;

$Time = \frac{Distance }{speed}\\\\Time = \frac{8}{15} \\\\Time = 0.533 \ hr$

The distance covered by the slower runner at this time is given by;

d₁ = u₁ x 0.533 hr

d₁ = 11.8 km/hr x 0.533 hr

d₁ = 6.29 km

Additional distance (x) the slower runner need to finish is given by;

6.29 km + x = 8km

x = 8 k m - 6.29 km

x = 1.71 km

Therefore, the slower runner is 1.71 km from the finish line when the fastest runner finishes the race.

5 0

4 years ago

A 910 kg car is approaching a loop-the-loop. The loop has a diameter of 50 m. Determine the minimum speed the car must have at t

scoundrel [369]

Answer:

The minimum speed the car must have at the top of the loop to not fall = 35 m/s

Explanation:

Anywhere else on the loop, the speed needed to keep the car in the loop is obtained from the force that keeps the body in circular motion around the loop which has to just match the force of gravity on the car. (Given that frictional force = 0)

mv²/r = mg

v² = gr = 9.8 × 25 = 245

v = 15.65 m/s

But at the top, the change in kinetic energy of the car must match the potential energy at the very top of the loop-the-loop

Change in kinetic energy = potential energy at the top

Change in kinetic energy = (mv₂² - mv₁²)/2

v₁ = velocity required to stay in the loop anywhere else = 15.65 m/s

v₂ = minimum velocity the car must have at the top of the loop to not fall

And potential energy at the top of the loop = mgh (where h = the diameter of the loop)

(mv₂² - mv₁²)/2 = mgh

(v₂² - v₁²) = 2gh

(v₂² - (15.65)²) = 2×9.8×50

v₂² - 245 = 980

v₂² = 1225

v₂ = 35 m/s

Hence, the minimum speed the car must have at the top of the loop to not fall = 35 m/s

4 0

3 years ago

Why houses have garage?

kykrilka [37]

To keep cars safe...............

3 0

3 years ago

Read 2 more answers

What acceleration reaches a car of Formla 1 of 650 Kg of mass whose engine starts with a force of 6000 N

PtichkaEL [24]

Hope this helps

650 x 6000= 3900000

4 0

3 years ago