Answer:
the force required to accelerate a 1,100kg car is 550N
Answer:
Proxima Centauri
Explanation:
U 2 can help me by marking as brainliest........
Thw question is not complete. The complete question is;
Charge of uniform linear density (6.7 nCim) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 1.6 m. a. 32 N/C b. 150 NC c 75 N/C d. 49 N/C e. 63 NC
Answer:
Option C: E = 75 N/C
Explanation:
We are given;
Uniform linear density; λ = 6.7 nC/m = 6.7 × 10^(-9) C/m
Distance on the y-axis; d = 1.6 m
Now, the formula for electric field with uniform linear density is given as;
E = λ/(2•π•r•ε_o)
Where;
E is electric field
λ is uniform linear density = 6.7 × 10^(-9) C/m
r is distance = 1.6m
ε_o is a constant = 8.85 × 10^(-12) C²/N.m²
Thus;
E = (6.7 × 10^(-9))/(2π × 1.6 × 8.85 × 10^(-12))
E = 75.31 N/C ≈ 75 N/C
Answer:
0.0109 m ≈ 10.9 mm
Explanation:
proton speed = 1 * 10^6 m/s
radius in which the proton moves = 20 m
<u>determine the radius of the circle in which an electron would move </u>
we will apply the formula for calculating the centripetal force for both proton and electron ( Lorentz force formula)
For proton :
Mp*V^2 / rp = qp *VB ∴ rp = Mp*V / qP*B ---------- ( 1 )
For electron:
re = Me*V/ qE * B -------- ( 2 )
Next: take the ratio of equations 1 and 2
re / rp = Me / Mp ( note: qE = qP = 1.6 * 10^-19 C )
∴ re ( radius of the electron orbit )
= ( Me / Mp ) rp
= ( 9.1 * 10^-31 / 1.67 * 10^-27 ) 20
= ( 5.45 * 10^-4 ) * 20
= 0.0109 m ≈ 10.9 mm
