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Nata [24]
3 years ago
12

an oscillator with a mass of 310 g and a period of 0.180 s has amplited that decreases by 20 % during each complete oscillation

what is the damping constant
Physics
1 answer:
Alex3 years ago
6 0

Answer:

1.241/s

Explanation:

From this question, we have the following information:

Mass of oscillator = 310

The time Period, t = 0.180

Decrease = 20 %

Amplitude of oscillation has the formula

A = Aoe^yt

1-20% = 80%

= 0.80

Y x 0.18 = -ln(0.8)

= 0.18Y = 0.2234

Y = 0.2233/0.18

Y = 1.241/s

The damping constant y is Therefore equal to 1.241s

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Inertia is directly proportional to mass.

What is Walter Lewin famous for?

Walter Hendrik Gustav Lewin (born January 29, 1936) is a Dutch astrophysicist and former professor of physics at the Massachusetts Institute of Technology.

Lewin earned his doctorate in nuclear physics in 1965 at the Delft University of Technology and was a member of MIT's physics faculty for 43 years beginning in 1966 until his retirement in 2009.

According to Walter Levin,

The concept of moment of inertia is demonstrated by rolling a series of cylinders down an inclined plane.

Inertia is the resistance of any physical object to a change in its velocity. This includes changes to the object's speed, or direction of motion. An aspect of this property is the tendency of objects to keep moving in a straight line at a constant speed when no forces act upon them.

By rolling a series of cylinders down on an inclined plane , he demonstrated that a cylinder have a smooth friction.

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7 0
2 years ago
Car a runs a red light and broadsides car b, which is stopped and waiting to make a left turn. car a has a mass of 1,800kg. car
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The law of conservation of momentum tells us that momentum is conserved, therefore total initial momentum should be equal to total final momentum. In this case, we can expressed this mathematically as:

mA vA + mB vB = m v

where, m is the mass in kg, v is the velocity in m/s

since m is the total mass, m = mA + mB, we can write the equation as:

mA vA + mB vB = (mA + mB) v

furthermore, car B was at a stop signal therefore vB = 0, hence

mA vA + 0 = (mA + mB) v

1800 (vA) = (1800 + 1500) (7.1 m/s)

<span>vA = 13.02 m/s</span>

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5 0
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Calculate the magnitude of the linear momentum for the following cases. (a) a proton with mass 1.67 10-27 kg, moving with a spee
abruzzese [7]

Answer:

<h2>8.0995×10^-21 kgms^-1</h2>

Explanation:

Mass of proton :

m_P=1.67\times 10^-^2^7\:kg\\

Speed of Proton:

v_P=4.85\times 10^6

Linear Momentum of a particle having mass (m) and velocity (v) :

-> p =m->v\:\:\: (1)

Magnitude of momentum :

p=mv\:\:\: (2)

Frome equation (2), magnitude of linear momentum of the proton :

p_P=m_P\:v_P\\\\p_P=1.67\times 10^-^2^7 \:kg\times4.85\times 10^6\:ms^-^1\\\\p_P= 8.0995\times 10^-^2^1\:kgms^-^1

7 0
3 years ago
Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of 20
Genrish500 [490]

Answer:

The time it can operate between chargins in minutes is

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I=\frac{1}{2}*m*r^2

I=\frac{1}{2}*500kg*(1.0m)^2

I=250 kg*m^2

K_R=\frac{1}{2}*250kg*m^2*(200\pi rad/s)^2

K_R=49.348x10^6J

b). The power average 0.8kW un range time can be find

P=\frac{K_R}{t'}

Solve to t'

t=\frac{K_R}{P}

t=\frac{49.348x10^6}{0.8x10^3w}=6168.5s

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3 years ago
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