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Vikki [24]
2 years ago
9

The magnetic field lines due to a straight, current-carrying wire are __________. The magnetic field lines due to a straight, cu

rrent-carrying wire ar
Physics
1 answer:
Rama09 [41]2 years ago
4 0

The magnetic field lines due to a straight, current-carrying wire are circular.

<u>Explanation:</u>

The concepts of Electromagnetism brought a new revolution to the science world. The idea is the source of many new modes of power and machines that reduces the manual work. Motors are the best example of machines that run on the concepts of electromagnetism. So the concept is that a current-carrying conductor induces a magnetic field in its nearby premise.

This magnetic field can perceive by the magnetic line of forces. Now, if we pour some iron dust around a current-carrying conductor, we'll see a concentric circular pattern around the straight wire whose centre will be at the conductor axis. The pattern of these magnetic lines of force may deflect with the variation of current in the wire but remain in the circular format.

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Answer:

The answer for the above statement is:

C. High-visibility clothing is important to wear in areas with moving vehicles.

because in bright clothes you are easier to see, so people driving can see you.

Explanation:

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A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud
IceJOKER [234]

Answer: 1.51 km

Explanation:

<u>Coulomb's Law:</u> The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or,   \vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

<u>Given:</u>

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

<u>So,</u>

\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514  \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}

Therefore, the separation between the two charges (r) = 1.51 km

3 0
2 years ago
b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 2.50 m/s
fomenos

Answer:

The horizontal distance is 0.64 m.

Explanation:

Initial velocity, u =2.5m/s

The maximum horizontal distance is

R = \frac{u^2}{2g}\\\\R = \frac{2.5\times 2.5}{9.8}\\\\R = 0.64 m

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3 years ago
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Which type of energy is transferred by convection currents?
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Explanation:
Moving particles transfer thermal energy through a fluid by forming convection currents.
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