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
Given:
A cylindrical container closed of both end has a radius of 7cm and height of 6cm.
Explanation:
A.) Find the total surface area of the container.
- A = 2πrh + 2πr²
- A = 2(3.14)(7)(6) + 2(3.14)(7 × 7)
- A = 263.76 + 307.72
- A = 571.48
B.) Find the volume of the container.
- V = πr²h
- V = (3.14)(7×7)(6)
- V = 923.16
Not sure huhuness.
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Hello
Here we must use the equation of motion
v^2 = u^2 + 2as; where v is final velocity, u is initial velocity, a is the acceleratoin and is the distance travelled.
We select this one because the time of collision is unknown to us.
We know the truck stopped so its final velocity is 0; thus v = 0.
Converting the initial velocity to SI units, we get 3.89 m/s.
The distance traveled, s, is 0.062 meters.
Inserting all of these values into the equation,
0 = (3.89)^2 + 2(a)(0.062)
and solving for a, we get a to be
-122.0 ms^(-2)
The negative sign indicates the acceleration is in the opposite direction to the initial motion, which means the truck decelerated. This is consistent with the given condition.
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Explanation:
