Answer:
3.75 m/s south
Explanation:
Momentum before collision = momentum after collision
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
Since the car and truck stick together, v₁ = v₂.
m₁ u₁ + m₂ u₂ = (m₁ + m₂) v
Given m₁ = 1500 kg, u₁ = -15 m/s, m₂ = 4500 kg, and u₂ = 0 m/s:
(1500 kg) (-15 m/s) + (4500 kg) (0 m/s) = (1500 kg + 4500 kg) v
-22500 kg m/s = 6000 kg v
v = -3.75 m/s
The final velocity is 3.75 m/s to the south.
Answer:
51.82
Explanation:
First of all, let's convert both vectors to cartesian coordinates:
Va = 36 < 53° = (36*cos(53), 36*sin(53))
Va = (21.67, 28.75)
Vb = 47 < 157° = (47*cos(157), 47*sin(157))
Vb = (-43.26, 18.36)
The sum of both vectors will be:
Va+Vb = (-21.59, 47.11) Now we will calculate the module of this vector:

Hello!
Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.

We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)
- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)
Both of these act in opposite directions. Let's use the equation for torque:

Doing the summation using their respective lever arms:


Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:

Now, let's solve for 'T'.

Plugging in the values:

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