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abruzzese [7]
3 years ago
15

Refer to a long, straight wire carrying constant current I. What can be concluded about the magnitude of the magnetic field at d

istance a from the wire?
Physics
1 answer:
sergij07 [2.7K]3 years ago
5 0

Answer:

<em>"the magnitude of the magnetic field at a point of distance a around a wire, carrying a constant current I, is inversely proportional to the distance a of the wire from that point"</em>

Explanation:

The magnitude of the magnetic field from a long straight wire (A approximately a finite length of wire at least for close points around the wire.) decreases with distance from the wire. It does not follow the inverse square rule as is the electric field from a point charge. We can then say that<em> "the magnitude of the magnetic field at a point of distance a around a wire, carrying a constant current I, is inversely proportional to the distance a of the wire from that point"</em>

From the Biot-Savart rule,

B = μI/2πR

where B is the magnitude of the magnetic field

I is the current through the wire

μ is the permeability of free space or vacuum

R is the distance between the point and the wire, in this case is = a

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Two ships leave a harbor at the same time, traveling on courses that have an angle of 110∘ between them. If the first ship trave
Allushta [10]

Answer:

49.07 miles

Explanation:

Angle between two ships = 110° = θ

First ship speed = 22 mph

Second ship speed = 34 mph

Distance covered by first ship after 1.2 hours = 22×1.2 = 26.4 miles = b

Distance covered by second ship after 1.2 hours = 34×1.2 = 40.8 miles = c

Here the angle between the two sides of a triangle is 110° so from the law of cosines we get

a² = b²+c²-2bc cosθ

⇒a² = 26.4²+40.8²-2×26.4×40.8 cos110

⇒a² = 2408.4

⇒a = 49.07 miles

6 0
3 years ago
11) An airplane is moving at a net force of zero. This means the airplane is
Elden [556K]

Answer:

D. moving slow at a constant speed and traveling along a straight line

3 0
3 years ago
All ball is thrown up with a vertical velocity of 54 m/s and a horizontal velocity of 39 m/s. Calculate how many seconds it will
VikaD [51]

5.5 s

Explanation:

The time it takes for the ball to reach its maximum height can be calculated using

v_y = v_{0y} - gt \Rightarrow t = \dfrac{v_{0y}}{g}

since v_y = 0 at the top of its trajectory. Plugging in the numbers,

t = \dfrac{(54\:\text{m/s})}{(9.8\:\text{m/s}^2)} = 5.5\:\text{s}

8 0
3 years ago
A particle of mass m collides with a second particle of mass m. Before the collision, the first particle is moving in the x-dire
oee [108]

Answer:

a) v, v

b) 2mv^2

c) Elastic collion

Explanation:

(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v).  From momentum conservation in x-direction

Here x, y represent direction.They are not variable. 1 and 2 represent before and after.

2vm=v1xm+v2xm, we find v1x=v.

From momentum conservation in y-direction

0 =v1ym+v2ym, we findv1y=v.

(b) By energy conservation principle

Before: K=1/2m(2v)^2=2mv^2.

After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2

(c) The collision is elastic

6 0
3 years ago
A 2kg object is tied to the end of a cord and whirled in a horizontal circle of radius 2 m. If the body makes three complete rev
Travka [436]

Answer:

a) 37.70 m/s

b)710.6 m/s²

Explanation:

Given that ;

Mass of object = 2 kg

Radius of the motion = 2m

Frequency of motion = 3 rev/s

The formula to apply is;

v= 2πrf   where v is linear speed

v = 2×π×2×3 =12π = 37.70 m/s

Centripetal acceleration is given as;

a= 4×π²×r×f²  

a= 4×π²×2×3²

a=710.6 m/s²

5 0
3 years ago
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