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Vsevolod [243]
3 years ago
5

In complete agreement​

Physics
1 answer:
valkas [14]3 years ago
8 0

Answer:

umm.....

Explanation:

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Two charges with a certain distance apart have an electrostatic force of 400 N acting
Doss [256]
As the charges’ distance increase, there is a weaker force of attraction between them hence the electrostatic force decreases as distance increases. It increase by 4 (times 4) so the force will decrease by 4 making the answer

=A (400 divided by 4 = 100)
3 0
2 years ago
Please help!!
baherus [9]

Answer:

Mass is the correct answer.

Explanation:

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What is the magnitude of the acceleration of an electron at a point where the electric field has magnitude 6377 n/c and is direc
shusha [124]
Use the magnitude acceleration formula .

4 0
3 years ago
A 5.5kg mass is pushed with a force of 31N across a table having μk of 0.350. Find how fast it will accelerate, taking friction
Ipatiy [6.2K]

Answer:

3.66m/s^2

Explanation:

First, we need to find the friction which is F=u*N.

After that, find the resultant force by substarcting the friction from the forward force.

Lastly,using the formula F=ma, substitute in the known values of F and m to find a

4 0
3 years ago
An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see
mash [69]

Answer:

(7.90 × 10⁻¹⁵) J

Explanation:

The electric force exerted on the elecrron by rhe electric field is given by

F = qE

where |q| = charge on the particle = (1.602 × 10⁻¹⁹) C

E = magnitude of the electric field = (2.9 × 10⁶) V/m or N/C

F = 1.602 × 10⁻¹⁹ × 2.9 × 10⁶ = (4.646 × 10⁻¹³) N

From Newton's first law of motion relation, we can obtain the acceleration this force confers on the electron

F = ma

m = mass of the electron = (9.11 × 10⁻³¹) kg

a = acceleration of the electron caused by the electric force = ?

(4.646 × 10⁻¹³) = (9.11 × 10⁻³¹) × a

a = (4.646 × 10⁻¹³)/(9.11 × 10⁻³¹)

a = (5.10 × 10¹⁷) m/s²

Now, using the equations of motion, we can obtain the velocity with which the electron reaches the positive plate

u = initial velocity of the electron = 0 m/s (since the electron was initially at rest)

v = final velocity of the electron = ?

a = acceleration of the electron = (5.10 × 10¹⁷) m/s²

y = distance covered by the electron = 1.7 cm = 0.017 m

v² = u² + 2ay

v² = 0² + 2(5.10 × 10¹⁷)(0.017)

v² = (1.734 × 10¹⁶)

v = 131,677,182.5 m/s = (1.32 × 10⁸) m/s

Kinetic energy with which the electron hits the positive plate = (1/2)(m)(v²) = (1/2)(9.11 × 10⁻³¹)(1.32 × 10⁸)² = (7.90 × 10⁻¹⁵) J

Hope this Helps!!!

3 0
2 years ago
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