Answer:
F= 2569.6 X 4.65 = 11,948.64
*Multiply the mass and the acceleration to find the force
Explanation:
Answer:
The x-component of the electric field at the origin = -11.74 N/C.
The y-component of the electric field at the origin = 97.41 N/C.
Explanation:
<u>Given:</u>
- Charge on first charged particle,
- Charge on the second charged particle,
- Position of the first charge =
- Position of the second charge =
The electric field at a point due to a charge at a point distance away is given by
where,
- = Coulomb's constant, having value
- = position vector of the point where the electric field is to be found with respect to the position of the charge .
- = unit vector along .
The electric field at the origin due to first charge is given by
is the position vector of the origin with respect to the position of the first charge.
Assuming, are the units vectors along x and y axes respectively.
Using these values,
The electric field at the origin due to the second charge is given by
is the position vector of the origin with respect to the position of the second charge.
Using these values,
The net electric field at the origin due to both the charges is given by
Thus,
x-component of the electric field at the origin = -11.74 N/C.
y-component of the electric field at the origin = 97.41 N/C.
Answer:
Magnetic field experienced = 4.5 × 10⁻⁴ T
Explanation:
The magnetic field around an infinite straight current-carrying wire at a distance r from the wire is given by
B = (μ₀I)/(2πr)
B = ?
I = 20 KA = 20000 A
r = 8.9 m
μ₀ = magnetic permeability = 1.257 × 10⁻⁶ T.m/A
B = (1.257 × 10⁻⁶ × 20000)/(2π×8.9) = 4.5 × 10⁻⁴ T
Changes. :) I think... Whats your question
?
Answer:
(a) The equivalent spring constant is 598.485 N/m
(b) The work done is 46.926 J
Explanation:
From Hooke's law of elasticity
K (spring constant) = F/e
F is the range of force exerted = 237 - 0 = 237 N
e is the extension of bowstring = 0.396 m
K = F/e = 237/0.396 = 598.485 N/m
Work done = 1/2 Fe = 1/2 × 237 × 0.396 = 46.926 J