Explain a situation in which you can have a positive velocity but a negative acceleration
1 answer:
You might be interested in
The answer is B
I used these equations then i putted it together.
Charge = number of ( electron or proton ) x charge of ( electron or proton )
Force = k x (q1 q2)/r²
I used these equations then i putted it together.
Charge = number of ( electron or proton ) x charge of ( electron or proton )
Force = k x (q1 q2)/r²
Answer:
Explanation:
Capacity of a parallel plate capacitor C = ε₀ A/ d
ε₀ is permittivity whose value is 8.85 x 10⁻¹² , A is plate area and d is distance between plate.
C =( 8.85 X10⁻¹² X 27 X 10⁻⁴ ) / 3 X 10⁻³
= 79.65 X 10⁻¹³ F.
potential diff between plate = Charge / capacity
= 4.8 X 10⁻⁹ / 79.65 X 10⁻¹³
= 601 V
Electric field = V/d
= 601 / 3 x 10⁻³
= 2 x 10⁵ N/C
Force on proton
= charge x electric field
1.6 x 10⁻¹⁹ x 2 x 10⁵
= 3.2 x 10⁻¹⁴
Acceleration a = force / mass
= 3.2 x 10⁻¹⁴ / 1.67 x 10⁻²⁷
= 1.9 x 10¹³ m s⁻²
Distance travelled by proton = 3 x 10⁻³
3 x 10⁻³ = 1/2 a t²
t =
t = 1.77 x 10⁻⁸ s