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3 years ago

What do you think happened to the energy you transferred to the notebook when you pushed it?

Physics

2 answers:

xeze [42]3 years ago

8 0

Answer:

It got transferred to kinetic energy

Explanation:

Scorpion4ik [409]3 years ago

7 0

Answer:

The energy changed from kinetic energy to heat energy. Friction causes heating.

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Scientists have discovered just over _____ different elements with unique properties.

Y_Kistochka [10]

100. (: hope this helps

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3 years ago

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A heavy boy and a lightweight girl are balanced on a mass-less seesaw. If they both move forward so that they are one-half their

Rom4ik [11]

Answer:

b) Nothing will happen, the sea saw will still be balanced.

Explanation:

b) Nothing will happen, the sea saw will still be balanced.

Reason:-

When two kids are balanced, the sum of torques on the seesaw will be zero.

if each kid, reduces their distances by half, then the torque of each kid will be half and the sum of torque of each on the seesaw will be zero.

Therefore the seesaw is balanced

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How can shorter string produce more fundamental frequency in two string of different length?

Nana76 [90]

Answer:

Shorter string produces more frequency in two different strings because the equation for frequency is velocity/wavelength , this means that a shorter string creates a shorter wavelength which essentially increases the total frequency produced

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3 years ago

Is ice cream a solid -or is it actually a solid?

blondinia [14]

It is a solid when is frozen and a liquid when it melts

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3 years ago

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

6 0

3 years ago