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algol [13]
2 years ago
10

Cho hệ thống thùng lắc có mô hình tại vị trí đang xét như hình vẽ

Physics
2 answers:
professor190 [17]2 years ago
4 0

Answer:

<h2>I can't understand this language!!!</h2>
Serga [27]2 years ago
4 0

Answer:

vdhdbdnnsnsbdhhshzbhshsbbsbd is not ask you to be able and r in the exam qq and

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When a force causes an object to move in the direction of the force blank occurs
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Its magnetic..........ig

7 0
3 years ago
A proton is 0.9 meters away from a 1.4 C charge. What is the magnitude of the electric force between the proton and the charge
Digiron [165]

Answer:

F = 2.49 x 10⁻⁹ N

Explanation:

The electrostatic force between two charged bodies is given by Colomb's Law:

F = \frac{kq_1q_2}{r^2}\\

where,

F = Electrostatic Force = ?

k = colomb's constant = 9 x 10⁹ N.m²/C²

q₁ = charge on proton = 1.6 x 10⁻¹⁹ C

q₂ = second charge = 1.4 C

r = distace between charges = 0.9 m

Therefore,

F = \frac{(9\ x\ 10^9\ N.m^2/C^2)(1.6\ x\ 10^{-19}\ C)(1.4\ C)}{(0.9\ m)^2}

<u>F = 2.49 x 10⁻⁹ N</u>

8 0
3 years ago
Read 2 more answers
An object is dropped from rest.
sesenic [268]

Taha xain malai ..........hhdd

3 0
3 years ago
A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in
koban [17]

Answer:

a) The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) The electric force is 2.0\times 10^{-6} newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

E = \frac{F_{e}}{q} (1)

Where:

E - Electric field, measured in newtons per coulomb.

F_{e} - Electric force, measured in newtons.

q - Electric charge, measured in coulombs.

If we know that F_{e} = 4.0\times 10^{-6}\,N and q = 2.0\times 10^{-8}\,C, then the electric field at that point is:

E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}

E = 2.0\times 10^{2}\,\frac{N}{C}

The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) If we know that E = 2.0\times 10^{2}\,\frac{N}{C} and q = 1.0\times 10^{-8}\,C, then the electric force is:

F_{e} = E\cdot q

F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)

F_{e} = 2.0\times 10^{-6}\,N

The electric force is 2.0\times 10^{-6} newtons.

7 0
3 years ago
Urgente!!!!! <br> Necesito ayuda con esto!!!!
ahrayia [7]
media.discordapp.net/attachments/782414373888458783/826224189828366377/video0.mp4
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3 years ago
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