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1 year ago

What is the relationship between speed and kinetic energy?

Physics

2 answers:

kobusy [5.1K]1 year ago

7 0

Answer:

Kinetic energy depends on the velocity of the object squared. This means that when the velocity of an object doubles, its kinetic energy quadruples.

Snowcat [4.5K]1 year ago

5 0

Answer:

The greater the speed of an object, the more kinetic energy it has.

Explanation:

Hope this helps! <3

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a crude approximation of voice production is to consider the breathing passages and mouth to be a resonating tube closed at one

prohojiy [21]

The fundamental frequency of the tube is 0.240 m long, by taking air temperature to be $37^o$ C is 367.42 Hz.

A standing wave is basically a superposition of two waves propagating opposite to each other having equal amplitude. This is the propagation in a tube.

The fundamental frequency in the tube is given by

$f=\frac{v_T}{4L}$

where, $v_T=v\sqrt{\frac{T}{273} }$

Since, T=37+273 K = 310 K

v = 331 m/s

$\therefore v_T=331\sqrt{\frac{310}{273} } = 352.72 \ m/s$

Using this, we get:

$f=\frac{352.72}{4(0.240)} \\f=367.42 \ Hz$

Hence, the fundamental frequency is 367.42 Hz.

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2 years ago

Which statement describes a characteristic of an experimental design that

Anarel [89]

Answer:

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3 years ago

Who yall think gonna win Seahawks or Vikings? Seahawks all the way

myrzilka [38]

Answer:

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Explanation:

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4 years ago

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A scientist wants to model an internal organ with connective tissue as a mass on a spring. The mass of the organ is 2.0 kg, and

vichka [17]

Answer:

Spring constant, k = 5483.11 N/m

Explanation:

It is given that,

Mass of the organ, m = 2 kg

The natural period of oscillation is, T = 0.12 s

Let k is the spring constant for the spring in the scientist's model. The period of oscillation is given by :

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

$k=\dfrac{4\pi^2 m}{T^2}$

$k=\dfrac{4\pi^2 \times 2\ kg}{(0.12\ s)^2}$

k = 5483.11 N/m

So, the spring constant for the spring in the scientist's model is 5483.11 N/m.

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3 years ago

A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca

svetoff [14.1K]

Answer:

$q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Explanation:

Given that $L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V$ , we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

$E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000 \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}$

#To find the particular solution:

$Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Hence the charge at any time, t is $q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

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4 years ago