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3 years ago

Can I have help? I REALLY need it! PLEASE!

Physics

1 answer:

valentina_108 [34]3 years ago

7 0

Parallel .............

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How does the gravitational force between two objects change if the distance

Anestetic [448]

Answer:

The Gravitational Force is reduced 4 times

Explanation:

The equation of Gravitational force follows:

F = (G*m1*m2)/r^2

Assume that G*m1*m2 = 1 and r = 1:

F = 1/1^2 = 1 N

Multiply the radius by 2

F = 1/2^2 = 1/4 N

So doubling the distance reduces the force 4 times.

7 0

3 years ago

Which dog has the most kinetic energy

Blababa [14]

The formula to find the kinetic energy is:

Ek= 1/2 × m × v^2

1. Ek= 1/2×15×3^2
= 67.5 J

2.Ek= 1/2×8×4^2
=64 J

3.Ek= 1/2×12×5^2
= 150 J

4.Ek= 1/2×10×6^2
= 180 J

So the fourth dog has the most kinetic energy.

4 0

3 years ago

a runner increases her speed from 3.1 m/s to 3.5 m/s during the last 15 seconds of her run, what was her acceleration during her

Rama09 [41]

Acceleration = velocity/time
A= 3.5m/s/15s
A= 0.23m/s^2

5 0

4 years ago

Read 2 more answers

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

1. Which of the following is not an acceptable Physical Education activity?

Dmitrij [34]

The answer is A

Explanation:
Vacuuming doesn’t involve a lot of physical movements.

3 0

3 years ago

Read 2 more answers