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julsineya [31]
3 years ago
10

Magnetic field lines, generated by a long straight wire carrying a current, are circles concentric with the wire and lie in plan

es perpendicular to the wire.
a. True
b. False
Physics
2 answers:
Aleksandr [31]3 years ago
6 0

Answer:

true

Explanation:

yes it is true.

To find the direction of magnetic field around a current carrying conductor, we use Maxwell;s right hand grip rule.

If we hold a current carrying straight conductor in our right hand such that the thumb indicates the direction of current in the wire then the curing of fingers give the direction of magnetic field around the conductor.

sammy [17]3 years ago
5 0

Answer:True

Explanation:

It is true that magnetic field lines around a current carrying wire is in the form of concentric circles and lie in the plane perpendicular to the wire.

This Fact can be Proven by placing the iron filling around the wire and they will be arranged like concentric circles and Magnetic field intensity diminishes as we go away from the wire.  

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Compared to its weight on Earth, a 10-kg object on the moon will weigh Question 9 options: less. the same amount. more.
Nostrana [21]

it will weigh less on the moon.

3 0
4 years ago
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An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in
valkas [14]

Answer:

Volume of the sample: approximately \rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}.

Average density of the sample: approximately \rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}.

Assumption:

  • \rm g = 9.81\; N \cdot kg^{-1}.
  • \rho(\text{water}) = \rm  1.000\times 10^{3}\; kg \cdot m^{-3}.
  • Volume of the cord is negligible.

Explanation:

<h3>Total volume of the sample</h3>

The size of the buoyant force is equal to \rm 17.50 - 11.20 = 6.30\; N.

That's also equal to the weight (weight, m \cdot g) of water that the object displaces. To find the mass of water displaced from its weight, divide weight with g.

\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg.

Assume that the density of water is \rho(\text{water}) = \rm  1.000\times 10^{3}\; kg \cdot m^{-3}. To the volume of water displaced from its mass, divide mass with density \rho(\text{water}).

\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}.

Assume that the volume of the cord is negligible. Since the sample is fully-immersed in water, its volume should be the same as the volume of water it displaces.

V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}.

<h3>Average Density of the sample</h3>

Average density is equal to mass over volume.

To find the mass of the sample from its weight, divide with g.

\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg.

The volume of the sample is found in the previous part.

Divide mass with volume to find the average density.

\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}.

3 0
4 years ago
______ is/are usually the best conductor/s. a. Metals c. Air b. Water d. Minerals Please select the best answer from the choices
nataly862011 [7]
Metals are the best conductors.
8 0
3 years ago
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A car traveling at 23 m/s starts to decelerate steadily. It comes to a complete stop in 5 seconds. What is its acceleration?
kykrilka [37]
We could determine the acceleration using this formula
\boxed{a= \dfrac{v_{1}-v_{0}}{t} }

Given from the question v₀ = 23 m/s, v₁ = 0 (the car stops), t = 5 s
plug in the numbers
a= \dfrac{v_{1}-v_{0}}{t}
a= \dfrac{0-23}{5}
a= \dfrac{-23}{5}
a = -4.6
The acceleration is -4.6 m/s²
8 0
3 years ago
Two protons (each with q = 1.60 x 10-19)
otez555 [7]

Answer:

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Explanation:

From the question given above, the following data were obtained:

Charge (q) of each protons = 1.6×10¯¹⁹ C

Distance apart (r) = 1×10¯¹⁵ m

Force (F) =?

NOTE: Electric constant (K) = 9×10⁹ Nm²/C²

The force exerted can be obtained as follow:

F = Kq₁q₂ / r²

F = 9×10⁹ × (1.6×10¯¹⁹)² / (1×10¯¹⁵)²

F = 9×10⁹ × 2.56×10¯³⁸ / 1×10¯³⁰

F = 2.304×10¯²⁸ / 1×10¯³⁰

F = 230.4 N

Therefore, the force exerted is 230.4 N

5 0
3 years ago
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