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erastova [34]
3 years ago
10

A 0.200 m wire is moved parallel to a 0.500 T

Physics
1 answer:
goldenfox [79]3 years ago
8 0

Answer:

The required emf moved across the wire is zero

Explanation:

For a moving charge particle, the magnetic force can be determined by using the formula;

\varepsilon = Bvlsin \theta

since the wire moves in parallel, the angle \theta between magnetic field and velocity = 0°

B = 0.500 T

v = 1.50 m/s

l = 0.200 m

∴

\varepsilon = (0.500  \ T )(1.50 \ m/s) \times (0.200 \ m)\times sin (0)

\varepsilon = 0.15\times sin (0)

\varepsilon = 0

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Mary and her younger brother Alex decide to ride the carousel at the State Fair. Mary sits on one of the horses in the outer sec
GaryK [48]

Mary and her younger brother Alex decide to ride the carousel at the State Fair, Mary's and Alex's  angular speed M and tangential speed vM is mathematically given as

Mary's and Alex's  angular speed=1.43

Tangential speed mary=3.22 m/s

Tangential speed alex =2.260m/s

<h3>What is Mary's and Alex's angular speed M and tangential speed vM?</h3>

Generally, the equation for angular speed is mathematically given as

w=2\pi /T\\\\Therefore\\\\w=2\pi/3.9

w = 1.61 rev/see 3.9

Centripetal acc mary = v^2/r

Centripetal acc mary  = w^2r

Centripetal acc mary = w^2x 2m

Centripetal acc. of Alex = w²x L.u

Therefore

\frac{(ac) mary }{(ac) plex}= 1.43

Hence

tang. speed V=Wr

tang. speed of mary = 1.61x2 = 3.22 m/s

tang. speed of Alex: 1.61X1·4 =2.260m/s

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8 0
2 years ago
Could anyone help with number 9?
Alisiya [41]
The answer would be A
3 0
3 years ago
At a distance 30 m from a jet engine, intensity of sound is 10 W/m^2. What is the intensity at a distance 180 m?
AleksandrR [38]

Answer:

I_{2}=0.27 W/m^2

Explanation:

Intensity is given by the expresion:

I_{2}=Io (\frac{r1}{r2} )^{2}

where:

Io = inicial intensity

r1= initial distance

r= final distance

I_{2}=10 W/m^2 (\frac{30m}{180m} )^{2}

I_{2}=0.27 W/m^2

5 0
3 years ago
A mechanic jacks up the front of a car to an angle of 8.0° with the horizontal in order to change the front tires. the car is 3.
SOVA2 [1]
<span>First draw a free-body diagram. Torque T = Force F x Distance d where force is the component of gravitational force g and d is the lever arm distance to the pivot point. Since the pivot point is at the back tire we subtract that from the length of the car resulting in d = 1.12 - 0.40 = 0.72 meters = d. We are interested in the perpendicular component of the force exerted on the car jack so use sin 8 degrees then T=1130 kg x 9.81 m/s^2 x sin(8 degrees) x0.72 m = 1,110.80 Newton-meters</span>
7 0
3 years ago
What is the electric field 4.0 m from the center of the terminal of a Van de Graaff with a 7.10 mC charge, noting that the field
Jobisdone [24]

Answer:

3984875 N/C

Explanation:

Applying,

E = kq/r².................. Equation 1

Where E = Electric field, q = charge, r = distance, k = coulomb's constant.

From the question,

Given: q = 7.10 mC = 0.0071 C, r = 4.0 m

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values into equation 1

E = 0.0071(8.98×10⁹)/4²

E = 3984875 N/C

Hence the electric field is 3984875 N/C

5 0
3 years ago
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