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Arisa [49]
3 years ago
9

Charge A is sitting in an electric field you know the following information:________

Physics
1 answer:
soldi70 [24.7K]3 years ago
6 0

Answer:

The equation that will relate all the given parameters, in other to calculate the potential energy of charge A is:

∆V = ∆U/q, ∆V is potential at charge A position, q is magnitude of charge A, ∆U will be made the subject of the relation, which is the Potential Energy of charge A. The notation "∆" show, the quantities have both in values and final values, in the electric field.(Change in Electric potential and potential energy, due to the effect of the field)

Explanation:

The potential energy of a charged particle (Charge A) in an electric field depends on the magnitude of the charge(Known as stated in the question). However, the potential energy per unit charge has a unique value at any point in the electric field.

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andrey2020 [161]

Explanation:

after 5 seconds, the velocity is (5s)(3m/s²) = 15m/s

The displacement after 5s is

x=vo + (1/2)at²

x = 0 + (1/2)(3m/s²)(5s)(5s)

x= 37.5 m

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3 years ago
¿Cuál es la frecuencia de una nota musical cuyo periodo es 0,005 s?
AlekseyPX

Answer:

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3 years ago
Each corner of a right-angled triangle is occupied by identical point charges "A", "B", and "C" respectively. Draw a sketch of t
NISA [10]

Answer:

Fnet = F√2

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Explanation:

A exerts a force F on B, and C exerts an equal force F on B perpendicular to that.  The net force can be found with Pythagorean theorem:

Fnet = √(F² + F²)

Fnet = F√2

The force between two charges particles is:

F = k q₁ q₂ / r²

where

k is Coulomb's constant, q₁ and q₂ are the charges, and r is the distance between the charges.

If we say the charge of each particle is q, then:

F = kq²/r²

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5 0
3 years ago
In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward
Vladimir79 [104]

Answer:

The answer is "1.01 \times 10^{-13}"

Explanation:

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KE=U=\frac{kqq'}{r}\\\\

Calculating the closest distance:

\to r=\frac{kqq'}{KE}\\\\

=\frac{k(2e)(79e)}{KE}\\\\=\frac{k(2)(79)e^2}{KE}\\\\=\frac{9.0\times 10^9 \ N \cdot \frac{m^2}{c}(2)(79)(1.6 \times10^{-19} \ C)^2}{(2.25\ meV) (\frac{1.6 \times 10^{-13} \ J}{1 \ MeV})}\\\\

=\frac{9.0\times 10^9 \times 2\times 79\times 1.6 \times10^{-19}\times 1.6 \times10^{-19} }{(2.25 \times 1.6 \times 10^{-13}) }\\\\=\frac{3,640.32\times 10^{-29}}{3.6 \times 10^{-13} }\\\\=\frac{3,640.32}{3.6} \times 10^{-16}\\\\=1011.2 \times 10^{-16}\\\\=1.01 \times 10^{-13}

5 0
3 years ago
Two metals of equal mass with different heat capacities are subjected to the same amount of heat. which undergoes the smallest c
liraira [26]
<span>So when two metals of equal mass but different heat capabilities are subjected to same heat quantity, the metal with higher heat capacity have the small temperature change. Heat supplied is determined as heat capacity of the metal times the change in temperature.</span>
8 0
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