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3 years ago

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Two long parallel wires placed side-by-side on a horizontal table carry identical size currents in opposite directions. The wire

on your right carries current toward you, and the wire on your left carries current away from you. From your point of view, the magnetic field at the point exactly midway between the two wiresA) points away from you.B) is zero.C) points toward you.D) points down.E) points up.

1 answer:

N76 [4]3 years ago

6 0

Answer:

D) Points down

Explanation:

We can find the direction of the magnetic field produced by each of the wire by using the right hand rule:

- the thumb must correspond to the direction of the current in the wire

- the other fingers wrap around the thumb and gives the direction of the magnetic field lines

For the wire on the right, we have:

- thumb (current): towards you

- other fingers (magnetic field): at the left of the wire, they point down

For the wire on the left, we have:

- thumb (current): away from you

- other fingers (magnetic field): at the right of the wire, they point down

So, both magnetic fields point down at the point halfway between the two wires, so the net field is also pointing down.

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A man jogs at a speed of 1.6 m/s. His dog waits 1.8 s and then takes off running at a speed of 3 m/s to catch the man. How far w

inessss [21]

Answer:

The dog catches up with the man 6.1714m later.

Explanation:

The first thing to take into account is the speed formula. It is $v=\frac{d}{t}$ , where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is $d=v\cdot t$

Now, the distance equation for the man would be:

$d_{man}=v_{man}\cdot t=1.6\cdot t$

The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.

$d_{dog}=v_{dog}\cdot (t-1.8)=3\cdot (t-1.8)$

For a better understanding, at t=1.8 the dog must be in d=0. Let's verify:

$d_{dog}=v_{dog}\cdot (1.8-1.8)=3\cdot (0)=0$

Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.

$d_{man}=d_{dog}$

$1.6\cdot t=3\cdot (t-1.8)$

$1.6\cdot t=3\cdot t-5.4$

$1.4\cdot t=5.4$

$t=\frac{5.4}{1.4}$

$t=3.8571s$

The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.

That value is used in the man's distance equation.

$d_{man}=1.6\cdot t=1.6\cdot (3.8571)$

$d_{man}=6.1714m$

Finally, the dog catches up with the man 6.1714m later.

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3 years ago

Where does current flows maximum in? series connection or parallel connection?

notsponge [240]

Current flow depends on other things in addition to the circuit configuration. If the SAME voltage is applied to some arrangement of the SAME components, the greatest current will occur when they are all in parallel.

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3 years ago

A 200kg ball on the end of string is swung in horizontal circle with radius of 0.5m . The ball makes revolution every 2second th

ANEK [815]

Answer: $\dfrac{\pi}{2} ms^{-1}$

Explanation:

Let $T$ be the time required to make one revolution.

Let $r$ be the radius of the circular path.

Let $d$ be the distance travelled by ball in one revolution.

As we know,the distance travelled in one revolution is the circumference of the circle.

So, $d=2\pi r$

Given, $d=0.5m\\T=2sec$

$d=2\times \pi \times 0.5=\pi m$

Speed of an object moving is circular path is define as the ratio of distance travelled in one revolution to the time taken by the object to complete one revolution.

Let $s$ be the speed of the ball.

$s=\frac{d}{T}=\frac{\pi }{2}ms^{-1}$

So,the speed of the ball is $\frac{\pi }{2}ms^{-1}$

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3 years ago

A 3.00 kg cart on a frictionless track is pulled by a string so that it accelerates at 2.00 m/s/s. what is the tension in the st

AnnZ [28]

Maybe the picture helps. The blue block represents the cart with a mass of 3 kg. The person(black block) is pulling the cart to the right with a force F so that the acceleration a is 2 m/s². According to Newton's 2nd law: F = m*a.

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3 years ago

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2. Xét một điểm nằm trên vành ngoài của lốp xe máy cách trục bánh xe môtô 25cm.

Rashid [163]

Answer:

b. Cho biết tốc độ của xe là 4m/s. Hãy tính tốc độ góc của điểm trên vành ngoài bánh xe.

Explanation:

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2 years ago