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ExtremeBDS [4]
3 years ago
11

How can you turn off a magnetic field produced by an electrical current?

Physics
2 answers:
sergij07 [2.7K]3 years ago
7 0

Answer:

d) By turning the electrical current off

Explanation:

this is the only one that fits this situation because all of the other answers help improve the magnetic field.

aleksklad [387]3 years ago
6 0

Answer:By turning the electrical current off

Explanation:Trust me I took the test

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What is the definition of Newton and could you leave an example of it too?
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What is the definition of newton
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If you were to move to the Canadian North Woods, what adaptations or behavioral changes would you make?
iragen [17]

Adaptation will mean taking action to minimize the negative effects of change. ... the use of new tools and techniques for decision-making, For example, projected increases in drought, fire, windstorms, and insect and disease outbreaks are expected to result in greater tree mortality. Fewer trees will reduce Canada’s timber supply, which in turn will affect the economic competitiveness of Canada’s forest industry. This would leave forestry-dependent communities vulnerable to job losses, closure of forestry processing facilities and an overall economic slump.

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) The centre of a mass, m, is at a distance, r, from the centre of a larger mass M. Assuming there are no other masses present,
ivanzaharov [21]

Answer:

U = - G m M / r

Explanation:

The gravitational potential energy is given by the expression

         U = - G m₁ m₂ / r

dodne G is the gravitational cosntnate (G = 6.67 10⁻¹¹¹), m and m are the mass of the bodies involved

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         U = - G m M / r

8 0
3 years ago
Which of the following is an example of static friction?
Tema [17]
The answer is A.......
5 0
3 years ago
A 16000 kg railroad car travels along on a level frictionless track with a constant speed of 23.0 m/s. A 5400 kg additional load
Alisiya [41]

Answer:

The speed of the car when load is dropped in it is 17.19 m/s.

Explanation:

It is given that,

Mass of the railroad car, m₁ = 16000 kg

Speed of the railroad car, v₁ = 23 m/s

Mass of additional load, m₂ = 5400 kg

The additional load is dropped onto the car. Let v will be its speed. On applying the conservation of momentum as :

m_1v_1=(m_1+m_2)v

v=\dfrac{m_1v_1}{m_1+m_2}

v=\dfrac{16000\ kg\times 23\ m/s}{(16000+5400)\ kg}

v = 17.19 m/s

So, the speed of the car when load is dropped in it is 17.19 m/s. Hence, this is the required solution.

6 0
3 years ago
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