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babunello [35]
3 years ago
10

A conventional current of 7 A runs clockwise in a circular loop of wire in the xy plane, with center at the origin and with radi

us 0.097 m. Another circular loop of wire lies in the same plane, with its center at the origin and with radius 0.03 m. How much conventional current must run counterclockwise in this smaller loop in order for the magnetic field at the origin to be zero?
Physics
1 answer:
Anni [7]3 years ago
4 0

Answer:

2.17 A

Explanation:

The magnetic field due to a circular current carrying coil is given by

B = k x 2i / r

For i = 7 A, r = 0.097 m, clockwise

B = k x 2 x 7 / 0.097 = 144.33 k (inwards)

The direction of magnetic field is given by the Maxwell's right hand thumb rule.

The magnetic field is same but in outwards direction as the current is in counter clockwise direction. Let the current be i.

Now, r = 0.03 m, B = 144.33 K, i = ?

B = k x 2i / r

144.33 K = K x 2 x i / 0.03

i = 2.17 A

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The United States and South Korean soccer teams are playing in the first round of the World Cup. An American kicks the ball, giv
garik1379 [7]

Answer:

Vf = 3.67 [m/s]

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f} ^{2} =v_{i} ^{2} -(2*a*x)

where:

Vf = final velocity [m/s]

Vi = initial velocity = 4.3 [m/s]

a = acceleration or desacceleration = 0.5 [m/s²]

x = distance = 5 [m]

Note: The negative sign in the above equation means that the velocity of the ball is decreasing (desacceleration).

Now replacing:

Vf² = (4.3)² - (2*0.5*5)

Vf² = 18.49 - 5

Vf² = 13.49

using the square root, we have.

Vf = 3.67 [m/s]

8 0
2 years ago
What element from the periodic table rhymes with extreme
jonny [76]
Halite or sulfur or gold or silver
8 0
3 years ago
A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed
liq [111]

Answer:

A) total time = 55.5 seconds

B) average velocity = 25.27 m/s

Explanation:

It starts from rest, so initial velocity, u = 0 m/s

We are given;

acceleration; a = 2 m/s²

Final velocity; v = 31 m/s

From Newton's first law of motion,

v = u + at

So, 31 = 0 + 2t

t = 31/2

t = 15.5 sec

We are told that, after this time of 15.5 sec, the car travels 35 sec at a constant speed and after that it takes 5 sec additional time to stop. Thus;

(a) Total time in which car is in motion = 15.5 + 35 +5 = 55.5 seconds

b)Total distance traveled during first 15.5 sec would be gotten from Newton's second equation of motion which is;

S = ut + ½at²

S1 = 0 + ½(2 * 15.5²)

S1 = 240.25 m

Distance traveled in 35 sec with with velocity of 31 m/sec is;

S2 = velocity x time

S2 = 35 × 31 = 1085 m

Now, for the final stage, final velocity (v) will now be 0 since the car comes to rest while initial velocity(u) will be 31 m/s.

From the first equation of motion,

a = (v - u)/t

a = (0 - 31)/5

a = -6.2 m/s²

So, distance travelled is;

S3 = ut + ½at²

S3 = (31 × 5) + ½(-6.2 × 5²)

S3 = 155 - 77.5

S3 = 77.5 m

So overall total distance = S1 + S2 + S3

Overall total distance = 240.25 + 1085 + 77.5 = 1402.75 m

Average velocity = total distance/total time

Average velocity = 1402.75/55.5 = 25.27 m/s

6 0
2 years ago
Two neutral metal spheres on wood stands are touching. A negatively charged rod is held directly above the top of the left spher
Paha777 [63]

Answer:

The right sphere is negatively charged, the left sphere is charged positively.

Explanation:

When a negatively charged rod is held above the top of left sphere, the rod will attract positive charges and repel negative charges. As the sphere are initially touching each other so positive charges from the both spheres will moves toward the rod. When we separate the spheres positive charges from right sphere have already moved toward the rod i.e. left sphere, creating a deficiency of positive charges in the right sphere and excessiveness of positive charges in left sphere , hence the right sphere will remain negatively charged and left sphere will remain positively charged.

4 0
3 years ago
While standing on a bridge 40.0 m above the ground, you drop a stone from rest. when the stone has fallen 3.80 m, you throw a se
Norma-Jean [14]
Given: distance 1 d₁ = 40 m;  distance 2 d₂ = 3.8 m   g = -9.8 m/s²

 Initial Velocity Vi = 0  Final Velocity of stone 2 is unknown = ?

Total distance dₓ = d₁ - d₂  = 40 m - 3.8 m = 36.2 m

Formula: a = Vf² -  Vi²/2d   derive for Final Velocity Vf
 
acceleration is now due to gravity, therefore a = g

Vf = √2gd   Vf = √2(9.8 m/s²)(36.2 m)

Vf = 26.64 m/s  

Reason: The second stone will still start from rest.


3 0
3 years ago
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