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2 years ago

10

The air going into the tyre was warmed up by the pumping. What effect will this have on the motion of Gad molecules in the air i

n the tyre?

1 answer:

Novosadov [1.4K]2 years ago

4 0

Answer:

they are constantly bouncing everywhere and creating preasure

Explanation:

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A speaker draws 2 amps of current when it is connected to a 16 volt source. What is the resistance of the speaker? Question 2 op

Oksana_A [137]

Did you ever find this answer

5 0

3 years ago

You hear a sound in the distance. Suddenly the sound gets deeper, decreasing in pitch. Which can you assume about the sound wave

oksian1 [2.3K]

Answer:

A. The wavelengths of the new sound waves are longer

Explanation:

This is the Doppler effect which can be best illustraded for the case of a siren of an ambulance approaching us having a greater frequency and getting lower in frequency and deeper as the ambulance passes us.

Since the wavelength is inversely proportional to the frequency it follows the wavelengths are longer when the frequency decreases lowering its pitch and getting deeper.

8 0

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

6 0

3 years ago

When temperature and pressure are applied to sedimentary and igneous rock?

spin [16.1K]

Answer:

Igneous rock

Explanation:

Igneous rocks are formed through the cooling and solidification of magma. It undergoes changes in temperature and pressure that causes it to cool, solidify, and crystallize.

6 0

3 years ago

What is induced current

Brrunno [24]

Answer:

A current can be induced in a conducting loop if it is exposed to a changing magnetic field. ... In other words, if the applied magnetic field is increasing, the current in the wire will flow in such a way that the magnetic field that it generates around the wire will decrease the applied magnetic field.

Explanation:

8 0

2 years ago