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Anastasy [175]
3 years ago
12

Identify the areas on the image where the force of repulsion is the least.​

Physics
2 answers:
astra-53 [7]3 years ago
6 0

Answer:

lower two magnets is the correct answer I believe

aalyn [17]3 years ago
4 0

Answer:

The central blue square in between the lower pair of magnet has the least force of repulsion.

Explanation:

We can explain this using the dual nature of magnets.

Each magnet must have two poles namely:

-North pole

-South pole

We assume that the magnetic lines of forces enters from south pole and leaves from the north pole.

When brought together, like poles repel each other while opposite poles attract each other.

In the picture, the lower two magnets have opposite poles facing each other, hence the force of repulsion is minimum there and the force of attraction is maximum.

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By what potential difference must a proton [m_0 = 1.67E-27 kg) be accelerated to have a wavelength lambda = 4.23E-12 m? By what
Vinil7 [7]

Explanation:

1. Mass of the proton, m_p=1.67\times 10^{-27}\ kg

Wavelength, \lambda_p=4.23\times 10^{-12}\ m

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

\lambda=\dfrac{h}{\sqrt{2m_pq_pV}}

V=\dfrac{h^2}{2q_pm_p\lambda^2}

V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 1.67\times 10^{-27}\times (4.23\times 10^{-12})^2}

V = 45.83 volts

2. Mass of the electron, m_p=9.1\times 10^{-31}\ kg

Wavelength, \lambda_p=4.23\times 10^{-12}\ m

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

\lambda=\dfrac{h}{\sqrt{2m_eq_eV}}

V=\dfrac{h^2}{2q_em_e\lambda^2}

V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 9.1\times 10^{-31}\times (4.23\times 10^{-12})^2}

V=6.92\times 10^{34}\ V

V = 84109.27 volt

Hence, this is the required solution.

7 0
3 years ago
What creates the changes in step 1,2,3 of the nitrogen cycle
Naya [18.7K]
1-fixation ( Bacteria Converts nitrogen to ammonium so plants can use it )
2-nitrification ( bacteria changes ammonium to nitrates and plants )
3 - Assimilation (plants absorb nitrates it is then used for Chlorophyll..)
6 0
3 years ago
How does momentum conservation work for a rocket firing its engines to speed up in deep space?
Mashutka [201]
Idkhhhhhhhhhhubvgbvcccc xzzz. Bcc bbb
3 0
3 years ago
What are successfulness of the Competition policy in South Africa?​
Mars2501 [29]

Answer:

The product choices along with its competitive prices were provided to the consumers. Practices such as horizontal collusion and resale price maintenance was declared unlawful in 1984. Prevention of monopoly growth was the aim of the competition policy act.

Explanation:

hope this helps you

6 0
3 years ago
An electric furnace runs 13 hours a day to heat a house during January (31 days). The heating element has a resistance of 7.2 an
liq [111]

Answer:

cost of running the furnace during January is $5619.62

Explanation:

given data

runs a day = 13 hours

January days = 31 days

resistance = 7.2 ohm

current = 16.7 A

cost of electricity = $0.10/kWh

to find out

cost of running the furnace during January

solution

first we get her power consumed by furnace that is

Power consumed = \frac{I^2}{R}  ........1

put here value we get

Power consumed = \frac{16.7^2}{7.2}

Power consumed = 38.7347 W

and

Power consumed by furnace in one hour is

Power consumed by furnace in one hour is = Power consumed × 3600

Power consumed by furnace in one hour is = 38.7347 × 3600  

Power consumed by furnace in one hour is 139.445kWh

and

Power consumed by furnace in the month of January is

Power consumed by furnace in the month of January = 139.445kWh × 13 hours × 31 days

Power consumed by furnace in the month of January = 56196.335 kWh

so

cost of running the furnace during January is = $0.10/kWh × 56196.335 kWh

cost of running the furnace during January is $5619.62

4 0
3 years ago
Read 2 more answers
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