Answer:
a) a = 4.9 m/s²
b) a = 1.5 m/s²
Explanation:
no friction
F = ma
gsinθ = ma
a = gsinθ
a = 9.8sin30
a = 4.9 m/s²
friction
gsinθ - μmgcosθ = ma
a = g(sinθ - μcosθ)
a = 9.8(sin30 - 0.4cos30)
a = 1.5051...
Answer:
6.78 X 10³ N/C
Explanation:
Electric field near a charged infinite plate
= surface charge density / 2ε₀
Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.
Field due to charge density of +95.0 nC/m2
E₁ = 95 x 10⁻⁹ / 2 ε₀
Field due to charge density of -25.0 nC/m2
E₂ = 25 x 10⁻⁹ / 2ε₀
Total field
E = E₁ + E₂
= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ / 2ε₀
= 6.78 X 10³ N/C
Answer:
Explanation:
Given that
As both charges are negative so there exist force of repulsion in direction as shown in figure.
Angle at which force F12 is acting is
Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction
Answer: By 47
Explanation: subtract 310 and 263
Answer:
7772.72N
Explanation:
When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.
Now which direction is the static friction, assume that it is pointing inward so
Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N
Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.
