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77julia77 [94]
2 years ago
5

What is damping??????​

Physics
2 answers:
Marianna [84]2 years ago
7 0
An influence upon an oscillatory system that has the effect of reducing/restriction or preventing its oscillations. Damping is produced by processes that dissipate the energy stored in the oscillation.
Ainat [17]2 years ago
3 0

Answer:

the picture is the definition

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Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n
jenyasd209 [6]

The statement about pointwise convergence follows because C is a complete metric space. If fn → f uniformly on S, then |fn(z) − fm(z)| ≤ |fn(z) − f(z)| + |f(z) − fm(z)|, hence {fn} is uniformly Cauchy. Conversely, if {fn} is uniformly Cauchy, it is pointwise Cauchy and therefore converges pointwise to a limit function f. If |fn(z)−fm(z)| ≤ ε for all n,m ≥ N and all z ∈ S, let m → ∞ to show that |fn(z)−f(z)|≤εforn≥N andallz∈S. Thusfn →f uniformlyonS.

2. This is immediate from (2.2.7).

3. We have f′(x) = (2/x3)e−1/x2 for x ̸= 0, and f′(0) = limh→0(1/h)e−1/h2 = 0. Since f(n)(x) is of the form pn(1/x)e−1/x2 for x ̸= 0, where pn is a polynomial, an induction argument shows that f(n)(0) = 0 for all n. If g is analytic on D(0,r) and g = f on (−r,r), then by (2.2.16), g(z) =

3 0
2 years ago
A 1200 kg car accelerates from 13m/s to 17m/s. find the change in momentum of the car?
lisabon 2012 [21]

Answer:

4800

Explanation:

The change in velocity of the car is 17-13=4m/s. Since the change in momentum is the mass multiplied by the change in velocity, the answer is 4*1200=4800. Hope this helps!

6 0
3 years ago
PLEASE HELP ASAP WILL GIVE BRAINLIEST!!!!
nydimaria [60]

Explanation:

Please mark me as the brainliest answer

7 0
2 years ago
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On a roller coaster, describe the kinetic and potential energy of the car at the top of the first hill, the bottom of the first
vaieri [72.5K]
The total amount of energy stays the same, but throughout the ride, the kinetic energy and the potential energy change, still adding up to the same number. At the top of the ride it has potential energy, and as it goes down the potential energy decreases and the kinetic energy increases. When it’s at the bottom of the first drop it has maxed out its kinetic energy, and minimized its potential energy. Friction slows down the car, and pushes on the cart with a force that is equal and opposite to the force being exerted in the track. The reason the track keeps going is because though it exerts and equal and opposite force the momentum of the objects is different, allowing the car to continue moving, however friction will slow it down until eventually it comes to a stop.
8 0
3 years ago
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Suppose a yo-yo has a center shaft that has a 0.230 cm radius and that its string is being pulled.
Fofino [41]

Answer:

Part a)

\alpha = 782.6 rad/s^2

Part B)

\omega = 587 rad/s

Part c)

a_t = 24.3 m/s^2

Explanation:

Part a)

As we know that

a = R \alpha

so we will have

a = 1.80 m/s^2

R = 0.230 cm

\alpha = \frac{a}{R}

\alpha = \frac{1.80}{0.230 \times 10^{-2}}

\alpha = 782.6 rad/s^2

Part B)

Angular speed of the yo-yo

\omega = \alpha t

so we have

\omega = 782.6 \times 0.750

\omega = 587 rad/s

Part c)

Tangential acceleration is given as

a_t = R \alpha

a_t = (3.10 \times 10^{-2})(782.6)

a_t = 24.3 m/s^2

6 0
3 years ago
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