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Novay_Z [31]
3 years ago
8

Vectors and have scalar product -9.00 and their vector product has magnitude 7.00.

Physics
1 answer:
sladkih [1.3K]3 years ago
6 0
<span>-(-(-)(-)(-10x))=-5 solve for x</span>
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An inductor of inductance 0.02H and capacitor of capacitance 2 microF are connected in series to an AC source of frequency 200/p
Ad libitum [116K]
C because it’s not a or B so 50/50 c or d and d is def not the answer so c
5 0
2 years ago
Electronic configuration of. <br>1)Fe. <br>2)Fe++ <br>3)Fe+++​
mezya [45]

Answer:

The electronic configuration of Fe2+ is 1s2 2s2 2p6 3s2 3p6 3d6 and Fe3+ is 1s2 2s2 2p6 3s2 3p6 3d5. Fe2+ contains 2 fewer electrons compared to the electronic configuration of Fe.

7 0
3 years ago
What can you say about the speed of the car?
iren2701 [21]
I believe the answer is C
5 0
3 years ago
When a negatively charged object moves in the opposite direction of an electric force field, the potential energy of the object
Simora [160]
The best and most correct answer among the choices provided by the question is decreases <span>.


</span>The potential energy of the object <span>decreases.</span>

Hope my answer would be a great help for you.    
If you have more questions feel free to ask here at Brainly.
3 0
3 years ago
Read 2 more answers
A positively charged particle is in the center of a parallel plate capacitor that has charge +/- Q on it's plates. Suppose the d
ivolga24 [154]

Answer:

the force remains constant if the charge does not change

Explanation:

In a capacitor the capacitance is given by

           C = ε₀ A / d

Where ε₀ is the permissiveness of emptiness, A is about the plates and d the distance between them.

The charge on the capacitor is given by the ratio

            Q = C ΔV

Let's apply these expressions to our problem, if the load remains constant

            C = Q / ΔV = ε₀ A / d

            ΔV / d = Q / ε₀ A

If the distance increases the capacitance should decrease, therefore if the charge is a constant the voltaje difference must increase

Now we can analyze the force on the test charge in the center of the capacitor

               ΔV = E d

               E= ΔV/d

               F = q E

              F = q ΔV / d

 Let's replace

          F = q Q /ε₀ A

From this expression we see that the force is constant since the voltage increase is compensated by increasing the distance, therefore the correct answer is that the force remains constant if the charge does not change

8 0
3 years ago
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