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MariettaO [177]
3 years ago
10

The small spheres that are moving through the circuit are the electric current. Current is the flow or movement of electrons. De

scribe how the current moves in the simulation.​
Physics
2 answers:
Grace [21]3 years ago
4 0

Answer:

When the circuit is connected properly, the current starts flowing from one end of the battery to the other end.

Explanation:

hope this helps!

patriot [66]3 years ago
3 0

Answer:

When the circuit is connected properly, the current starts flowing from one end of the battery to the other end.

Explanation:

Edmentum answer.

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Two particles, each of charge Q, are fixed at opposite corners of a square that lies in the plane of the page. A positive test c
amid [387]

Answer:

The magnitude of the net force is √2F.

Explanation:

Since the two particles have the same charge Q, they exert the same force on the test charge; both attractive or repulsive. So, the angle between the two forces is 90° in any case. Now, as we know the magnitude of these forces and that they form a 90° angle, we can use the Pythagorean Theorem to calculate the magnitude of the resultant net force:

F_N=\sqrt{F^{2}+F^{2}}\\\\F_N=\sqrt{2F^{2}}\\\\F_N=\sqrt{2}F

Then, it means that the net force acting on the test charge has a magnitude of √2F.

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3 years ago
Is an electrical current which comes from a battery
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3 years ago
A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s
Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g=4.67\times 10^{-3} kg

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1,m_1=1177g=1.177kg

Mass of block 2,m_2=1626 g=1.626 kg

Velocity of block 1,v_1=0.681m/s

(a)

Let velocity of the second block  after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

mv=m_1v_1+(m+m_2)v_2

4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2

1.66719=0.801537+1.63067v_2

1.66719-0.801537=1.63067v_2

0.865653=1.63067v_2

v_2=\frac{0.865653}{1.63067}

v_2=0.531m/s

Hence, the  velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

K_i=\frac{1}{2}mv^2

k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)

k_i=297.59 J

Final kinetic energy after collision

K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2

K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2

K_f=0.5028 J

Now, he ratio of the total kinetic energy after the collision to that before the collision

=\frac{k_f}{k_i}=\frac{0.5028}{297.59}

=0.00169

5 0
3 years ago
What are the ending materials produced from cell respiration
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Carbon dioxide and water :) hope this helped!
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2 years ago
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Compounds are formed as a result of ____
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Compounds are formed as a result of elements that are joined and held together by strong forces called chemical bonds.
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