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jek_recluse [69]
3 years ago
10

A metal rod has a moves with a constant velocity of 40 cm/s along two parallel metal rails through a magnetic field of 0.575 T.

If the rails are separated by 20 cm, what is the power generated by the rod? O 341 x 1ow O 6.98 x 103 w O 9.00x 10sw O 8.10 x 105w
Physics
1 answer:
love history [14]3 years ago
3 0

Answer:

2.12/R mW

Explanation:

The electrical power, P generated by the rod is

P = B²L²v²/R where B = magnetic field = 0.575 T, L = length of metal rod = separation of metal rails = 20 cm = 0.2 m, v = velocity of metal rod = 40 cm/s = 0.4 m/s and R = resistance of rod = ?

So, the induced emf on the conductor is

E = BLv

= 0.575 T × 0.2 m × 0.4 m/s

= 0.046 V

= 46 mV

The electrical power, P generated by the rod is

P = B²L²v²/R

=  B²L²v²/R

So, P = (0.575 T)² × (0.2 m)² × (0.4 m/s)²

= 0.002116/R W

= 2.12/R mW

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N experiment is performed in deep space with two uniform spheres, one with mass 27.0 and the other with mass 107.0 . They have e
Reptile [31]

Answer:

Explanation:

Apply the law of conservation of energy

KE_i+PE_i=KE_f+PE_f

Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)

from the law of conservation of the linear momentum

m_1v_1=m_2v_2

Therefore,

Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)

=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]

v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]

Substitute the values in the above result

v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]

=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s

B)  the speed of the sphere with mass 107.0 kg is

v_2=\frac{m_1v_1}{m_2}

=[\frac{27}{107} ](1.2664 \times 10^-^5)\\\\=3.195\times 10^-^6m/s

C)  the magnitude of the relative velocity with which one sphere is

v_r=v_1+v_2\\\\=1.2664\times 10^-^5+3.195\times10^-^6\\\\=15.859\times10^-^6m/s

D) the distance of the centre is proportional to the acceleration

\frac{x_1}{x_2} =\frac{a_1}{a_2} \\\\=\frac{m_2}{m_1} \\\\=3.962

Thus,

x_1=3.962x_2

and

x_2=0.252x_1

When the sphere make contact with eachother

Therefore,

x_1+x_2+2r=41\\x_1+0,252x_1+2r=41\\1.252x_1+2r=41\\x_1=32.747-1.597r

And

x_1+x_2+2r=41\\3.962x_2+x_2+2r+41\\4.962x_2+2r=41\\x_2=8.262-0.403r

The point of contact of the sphere is

32.747-1.597r=8.262-0.403r\\\\r=\frac{24.485}{1.194} \\\\=20.506m

3 0
3 years ago
How would the model change as the atom forms bonds? The third shell would have eight electrons after the atom gains seven electr
atroni [7]

Answer:

The third shell would be empty, so the eight electrons on the second level would be the outermost after the atom lost one electron

Explanation:

When an atom is bonded with other atoms, a more stable configuration must be reached, which is why the energy of the molecule is less than the energy of the individual atoms, for this to happen in general, electrons are shared or lost and gained in each atom, depending on the electronegative of the same.

If we analyze an atom within the molecule, its last shell is full, in the case of atoms with few electrons in this shell, they are lost and in the case of many electors in this shell, it gains electrons to have eight (8) in total.

When reviewing the different answers, the correct one is:

* The third shell would be empty, so the eight electrons on the second level would be the outermost after the atom lost one electron

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2 years ago
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A hill that has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction. At
Andreas93 [3]

The angle of incline of the hill above the horizontal is 8.81°.

Since the hill has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction.

<h3>Tangent of the angle of the incline of the hill,</h3>

The tangent of the angle of the incline of the hill, Ф is

tanФ = vertical rise/horizontal distance = grade of hill

Now, the vertical rise = 15.5 m and the horizontal distance = 100.0 m

So, substituting the values of the variables into the equation, we have

tanФ = vertical rise/horizontal run

tanФ = 15.5 m/100.0 m

tanФ = 0.155

<h3>Angle of incline of the hill</h3>

Taking inverse tan of both sides, we have

Ф = tan⁻¹(0.155)

Ф = 8.81°

So, the angle of incline of the hill above the horizontal is 8.81°.

Learn more about angle of incline of a hill here:

brainly.com/question/10056962

6 0
2 years ago
Which of the following is the best name for CaF2?
nikdorinn [45]
Its just Calcium Flouride because as you can see it is an Ionic bond (Ca is  metal and fl is a gas usually)
6 0
3 years ago
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A crate is placed on an adjustable, incline board. the coefficient of static friction between the crate and the board is 0.29.
sasho [114]

Let the angle be Θ (theta)

Let the mass of the crate be m.

a) When the crate just begins to slip. At that moment the net force will be equal to zero and the static friction will be at the maximum vale.

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μ (coefficient of static friction) = 0.29

Static friction = μN = μmg CosΘ

Now, along the ramp, the equation of net force will be:

mg SinΘ - μmg CosΘ = 0

mg SinΘ = μmg CosΘ

tan Θ = μ

tan Θ = 0.29

Θ = 16.17°

b) Let the acceleration be a.

Coefficient of kinetic friction = μ = 0.26

Now, the equation of net force will be:

mg sinΘ - μ mg CosΘ = ma

a = g SinΘ - μg CosΘ

Plugging the values

a = 9.8 × 0.278 - 0.26 × 9.8 × 0.96

a = 2.7244 - 2.44608

a = 0.278 m/s^2

Hence, the acceleration is 0.278 m/s^2

7 0
3 years ago
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