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2 years ago

Which is an example of a non-contact force?

Physics

2 answers:

nika2105 [10]2 years ago

7 0

Answer:

Explanation:

Because your not physically touching the item getting pulled towards the magnet. hope this helps

kotykmax [81]2 years ago

4 0

Answer:

B - magnet pulls an item toward it.

Explanation:

It's the answer trust bro

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Which options correctly describe the velocity of the object represented in the graph?

kirza4 [7]

There is no graph ...

7 0

2 years ago

Definition of graph?

xeze [42]

<u>Answer:</u>

1. A graph is defined as <em>" A Diagram represents a system of connections or interrelations among two or more things by a number of different dots, lines etc".</em>

2. In simple words <em>"Graph is a representation of any object or a physical structure by dots, lines, etc.</em>

6 0

3 years ago

A 4kg object has a momentum of 12 kg*m/s, what is the objects velocity?

Bezzdna [24]

Momentum = mass x velocity
12 = 4 x v | ÷ both sides by 4
12 ÷ 4 =v
v= 3 m/s

8 0

2 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

If a point charge is located at the center of a cylinder and the electric flux leaving one end of the cylinder is 20% of the tot

zheka24 [161]

The portion of the flux leaves the curved surface of the cylinder is 60%.

<h3 /><h3>What are electrons?</h3>

The electrons are the spinning objects around the nucleus of the atom of the element in an orbit.

If a point charge is located at the center of a cylinder and the electric flux leaving one end of the cylinder is 20% of the total flux leaving the cylinder.

If 20% of the flux leave from one end, then another 20% will leave from another end.

So, the net flux through curved surface is

100 -20 -20 = 60%

Thus, the total flux leaves the curved surface of the cylinder is 60%

Learn more about electrons.

brainly.com/question/1255220

#SPJ1

5 0

1 year ago