Speed is D. Distance per time.
Answer:
The acceleration is:
Explanation:
To answer this, we need to recall Newton's Second Law of motion, that states that an object of mass m would accelerate (change its state of uniform motion) proportional to the force (F) that is applied , and exemplified by the following equation:
From here, and using the given values for mass (m = 3 kg) and force (F = 9 N), we can derive the value of the acceleration as shown below (notice that since all quantities are given in SI units, the resulting acceleration will also be in Si units ():
Answer:
Explanation:
We know that, the force responsible for circulating in circular path is the centripetal force given by the force on charged particle due to magnetic field.
Here the charge is antiproton is
p = -1.6 * 10⁻¹⁹C
the speed of proton is given by 1.5 * 10 ⁴ m/s
the magnetic field is B = 4.5 * 10⁻³T
we have force due to magnetic field is equal to centripetal force
Bqv = mv² / r
Bq = mv / r
The diameter d of the vacuum chamber have to allow these antiprotons to circulate without touching the walls is
d = 2r
d = 2 * 3.479
d = 6.958
d ≅ 7cm
The electric potential at the origin of the xy coordinate system is negative infinity
<h3>What is the electric field due to the 4.0 μC charge?</h3>
The electric field due to the 4.0 μC charge is E = kq/r² where
- k = electric constant = 9.0 × 10 Nm²/C²,
- q = 4.0 μC = 4.0 × 10 C and
- r = distance of charge from origin = x₁ - 0 = 2.0 m - 0 m = 2.0 m
<h3>What is the electric field due to the -4.0 μC charge?</h3>
The electric field due to the -4.0 μC charge is E = kq'/r² where
- k = electric constant = 9.0 × 10 Nm²/C²,
- q' = -4.0 μC = -4.0 × 10 C and
- r = distance of charge from origin = 0 - x₂ = 0 - (-2.0 m) = 0 m + 2.0 m = 2.0 m
Since both electric fields are equal in magnitude and directed along the negative x-axis, the net electric field at the origin is
E" = E + E'
= -2E
= -2kq/r²
<h3>What is the electric potential at the origin?</h3>
So, the electric potential at the origin is V = -∫₂⁰E".dr
= -∫₂⁰-2kq/r².dr
Since E and dr = dx are parallel and r = x, we have
= -∫₂⁰-2kqdxcos0/x²
= 2kq∫₂⁰dx/x²
= 2kq[-1/x]₂⁰
= -2kq[1/x]₂⁰
= -2kq[1/0 - 1/2]
= -2kq[∞ - 1/2]
= -2kq[∞]
= -∞
So, the electric potential at the origin of the xy coordinate system is negative infinity
Learn more about electric potential here:
brainly.com/question/26978411
#SPJ11