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Andrews [41]
3 years ago
9

Which contribution did Kepler make to viewing the solar system?

Physics
1 answer:
amm18123 years ago
8 0
He proposed the sun-centered model of the solar system.
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What force must act on a 50.0-kg mass to give it an ancceleration of 0.30 m/s^2?
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15.0 N

Explanation:

see pic

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Solenoid is another word for
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What type of issue does the effects manufacturer need
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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

6 0
3 years ago
As a space shuttle moves through the dilute ionized gas of Earth's ionosphere, the shuttle's potential is typically changed by -
postnew [5]

Answer:

-2.3 × 10^-9 Coulombs(C).

Explanation:

So, we are given the following data or information or parameters that is going to help us to solve the problem effectively and efficiently;

=> " the shuttle's potential is typically changed by -1.4 V during one revolution. "

=> " Assuming the shuttle is a conducting sphere of radius 15 m".

So, in order to estimate the value for the charge we will be making use of the equation below:

Charge, C =( radius × voltage or potential difference) ÷ Coulomb's law constant.

Note that the value of Coulomb's law constant = 9 x 10^9 Nm^2 / C^2.

So, charge = { 15 × (- 1.4)} / 9 x 10^9 Nm^2 / C^2.

= -2.3 × 10^-9 Coulombs(C).

4 0
3 years ago
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