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anzhelika [568]
3 years ago
9

A segment of wire carries a current of 25 A along the x axis from x = −2.0 m to x = 0 and then along the y axis from y = 0 to y

= 3.0 m. In this region of space, the magnetic field is equal to 40 mT in the positive z direction. What is the magnitude of the force on this segment of wire?

Physics
1 answer:
julsineya [31]3 years ago
3 0

Answer: F = 2N

Explanation: If a current i is flowing in a wire of length L lying in a region of magnetic field B, then the magnetic force acting on the wire is given by

F = BIL

Please find the attached file for the solution

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Which order shows decreasing momentum?
klio [65]

Answer:

Explanation: Decreasing in velocity

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A 1451 kg car is traveling at 48.0 km/h. How much kinetic energy does it possess? K.E. =
jarptica [38.1K]

           Kinetic energy  =  (1/2) (mass) (speed)²

BUT . . . in order to use this equation just the way it's written,
the speed has to be in meters per second.  So we'll have to
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        KE  =  (1/2) · (1,451 kg) · (48 km/hr)² · (1000 m/km)² · (1 hr/3,600 sec)²

               =  (725.5) · (48 · 1000 · 1 / 3,600)²  (kg) · (km·m·hr / hr·km·sec)²

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8 0
3 years ago
A 2000-kg car moving with a speed of 20 m/s collides with and sticks to a 1500-kg car at rest at a stop sign. Show that because
amid [387]

Answer:

13.33m/s

Explanation:

Given data

m1= 2000kg

u1= 20m/s

m2= 1500kg

u2= 0m/s

v1= 10m/s

Required

The speed of the sticks

We know that  from the expression for the conservation of momentum

m1u1+m2u2= m1v1+m2v2

2000*20+1500*0=2000*10+1500*v2

40000=20000+1500v2

collect like terms

40000-20000= 1500v2

20000= 1500v2

v2= 20000/1500

v2= 13.33 m/s

Hence the velocity of the sticks is 13.33m/s

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A girl on a spinning amusements park is 12m from the center of the ride and has a centripetal acceleration of 17 m/s^2. What is
Tanya [424]

Answer: 14.28 m/s

Explanation:

Assuming the girl is spinning with <u>uniform circular motion</u>, her centripetal acceleration a_{c} is given by the following equation:  

a_{c}=\frac{V^{2}}{r} (1)

Where:  

a_{c}=17 m/s^{2} is the <u>centripetal acceleration</u>

V is the<u> tangential speed</u>

r=12 m is the <u>radius</u> of the circle

Isolating V from (1):

V=\sqrt{a_{c}r} (2)

V=\sqrt{(17 m/s^{2})(12 m)}

<u />

Finally:

V=14.28 m/s This is the girl's tangential speed

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