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lyudmila [28]
3 years ago
11

¿Qué es una vibración?

Physics
1 answer:
labwork [276]3 years ago
8 0

Answer:

A vibration or oscillation is a periodically repeated reversal of the direction of movement.

Explanation:

Une vibration ou une oscillation est une inversion répétée périodiquement de la direction du mouvement.

Here is just another definition.

A vibration is periodic back-and-forth motion centerd around it's equilibrium.

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A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
aivan3 [116]

Answer:

Explanation:

Given that,

Initial angular velocity is 0

ωo=0rad/s

It has angular velocity of 11rev/sec

ωi=11rev/sec

1rev=2πrad

Then, wi=11rev/sec ×2πrad

wi=22πrad/sec

And after 30 revolution

θ=30revolution

θ=30×2πrad

θ=60πrad

Final angular velocity is

ωf=18rev/sec

ωf=18×2πrad/sec

ωf=36πrad/sec

a. Angular acceleration(α)

Then, angular acceleration is given as

wf²=wi²+2αθ

(36π)²=(22π)²+2α×60π

(36π)²-(22π)²=120πα

Then, 120πα = 8014.119

α=8014.119/120π

α=21.26 rad/s²

Let. convert to revolution /sec²

α=21.26/2π

α=3.38rev/sec

b. Time Taken to complete 30revolution

θ=60πrad

∆θ= ½(wf+wi)•t

60π=½(36π+22π)t

60π×2=58πt

Then, t=120π/58π

t=2.07seconds

c. Time to reach 11rev/sec

wf=wo+αt

22π=0+21.26t

22π=21.26t

Then, t=22π/21.26

t=3.251seconds

d. Number of revolution to get to 11rev/s

∆θ= ½(wf+wo)•t

∆θ= ½(0+11)•3.251

∆θ= ½(11)•3.251

∆θ= 17.88rev.

5 0
3 years ago
Read 2 more answers
An object is released from height of 17m. <br> The object will hit the ground approximately in
zzz [600]

\text{Given that,}\\\\\text{Height, h = 17 m}\\\\\\\text{We know that,}\\\\h = v_0t + \dfrac 12  gt^2\\\\\implies h = \dfrac 12 gt^2\\\\\implies t^2 = \dfrac{2h}g\\\\\implies t =\sqrt{\dfrac{2h}g} = \sqrt{\dfrac{2(17)}{9.81}} = 1.87 ~ \text{sec}

5 0
2 years ago
A material through which electronics do NOT easily flow is
ziro4ka [17]
Resistor. The answer is a resistor. They do not transmit electricity. 
6 0
3 years ago
A car going at v = 29.7 m/s (67 mph) rounds a curve of radius R = 50.0 m, where the road is banked at an angle of θ = 30.0°. Wha
Nikolay [14]

Answer:

μ = 0.6

Explanation:

given,

speed of car = 29.7 m/s

Radius of curve = 50 m

θ = 30.0°

minimum static friction = ?

now,

writing all the forces acting along y-direction

N cos θ - f sinθ = mg

N cos θ -μN sinθ = mg

N = \dfrac{m g}{cos\theta-\mu sin \theta}

now, writing the forces acting along x- direction

N sin θ + f cos θ = F_{net}

N cos θ + μN sinθ = F_{net}

\dfrac{m g}{cos\theta-\mu sin \theta}(cos \theta + \mu sin\theta)=F_{net}

taking cos θ from nominator and denominator

F_{net} =\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{mv^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{v^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}g

\mu=\dfrac{v^2 -r g tan\theta}{v^2tan\theta + r g}

now, inserting all the given values

\mu=\dfrac{29.7^2 -50 \times 9.8tan 30^0}{29.7^2\times tan 30^0 +50 \times 9.8}

μ = 0.6

7 0
3 years ago
PLEASE HELP.. I WOULD BE SO HAPPY!! ILL MAKE YOLO THE BRAINLIEST....
madreJ [45]

Answer:

YOLOOOoo

Explanation:

6 0
3 years ago
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