Answer:
d. It is equal to the component of the gravitational force acting down the ramp.
Explanation:
The stationary crate is inclined at an angle with the horizontal. The Recall, Frictional Force is any Force that opposes motion.
Because the Force of Friction that is opposing the motion of the crate along the inclination side.
Therefore this Frictional force is balanced or equal to the force that is driving the inclined force.
Hence Frictional Force is equal to the Gravitational Force that is acting in the ramp, that is why the crate is stationery.
The answer to this question is a) sulfur
Answer:
a) a = - 0.106 m/s^2 (←)
b) T = 12215.1064 N
Explanation:
If
F₁ = 9*1350 N = 12150 N (→)
F₂ = 9*1365 N = 12285 N (←)
∑Fx = M*a = (M₁ +M₂)*a (→)
F₁ - F₂ = (M₁ +M₂)*a
→ a = (F₁ - F₂) / (M₁ +M₂ ) = (12150-12285)N/(9*68+9*73)Kg
→ a = - 0.106 m/s^2 (←)
(b) What is the tension in the section of rope between the teams?
If we apply ∑Fx = M*a for the team 1
F₁ - T = - M₁*a ⇒ T = F₁ + M₁*a
⇒ T = 12150 N + (9 * 68 Kg)*(0.106 m/s^2)
⇒ T = 12215.1064 N
If we choose the team 2 we get
- F₂ + T = - M₂*a ⇒ T = F₂ - M₂*a
⇒ T = 12285 N - (9 * 73 Kg)*(0.106 m/s^2)
⇒ T = 12215.1064 N
Answer:
a) x = 0.200 m
b)E = 3.84*10^{-4} N/C
Explanation:
DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m
by relation for electric field we have following relation
according to question E = 0
FROM FIGURE
x is the distance from left point charge where electric field is zero
solving for x we get
x = 0.200 m
b)electric field at half way mean x =0.25
E = 3.84*10^{-4} N/C
