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aalyn [17]
3 years ago
6

Is it important to constantly stay connected for you?

Physics
1 answer:
igomit [66]3 years ago
3 0

Answer:

no

Explanation:

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When a wire is made smaller, the resistance increases. Which happens to the electric current?
Free_Kalibri [48]
Well, you can pretty much see it with this formula :

I = V/R

If the resistance (R) is increased, the amount of  I will be increased.

so the answer would be : A. The current decreases

hope this helps


7 0
4 years ago
Read 2 more answers
. If an 8 N force pulled a 3.0 kg object to the left and a 5 N force pushed it to the right,
maxonik [38]

Answer:

8-5=3

The bigger force is to the left

3N to the left

Explanation:

7 0
1 year ago
Hi people; there's a question that has been so confusing for me and a friend of mine.
Alborosie

Answer:

The correct answer is: 0°C + 0°C = 32°F

Explanation:

We need to remember the conversion equation from Celsius to Fahrenheit:

y^{\circ}F=(x^{\circ}C * \frac{9}{5})+32

In our case x = 0, then y will be:

(0^{\circ}C * \frac{9}{5})+32=32

y=32^{\circ}F

Now 0°C + 0°C is just 0°C because if we add a body at a certain temperature to another body with the same temperature the total temperature will the same.

Then, knowing that 0°C = 32°F we can conclude that:

0^{\circ}C+0^{\circ}C=32^{\circ}F

I hope it helps you!              

7 0
3 years ago
What is the first and second law of thermodynamics
ElenaW [278]
<span>The first law, also known as Law of Conservation of Energy, states that energy cannot be created or destroyed in an isolated system. The second law of thermodynamics states that the entropy of any isolated system always increases. hope this helps you!</span>
6 0
3 years ago
Read 2 more answers
A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences
NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, q=3\times 10^{-6}\ C

Mass of charge particle, m=2\times 10^{-6}\ C

Speed, v=5\times 10^{6}\ m/s

Acceleration, a=3\times 10^{4}\ m/s^2

We need to find the minimum magnetic field that would produce such an acceleration. So,

ma=qvB\ sin\theta

For minimum magnetic field,

ma=qvB

B=\dfrac{ma}{qv}

B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0
3 years ago
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