Hey there!
Seems like you're looking for the size and direction to the final velocity of the two cars. To find it, you must solve it like this.
0.4 kg(3 m/s) + 0.8kg(–2 m/s) = 1.2 kg m/s -1.6 kg m/s = –0.4 kg m/s
–0.4 kg m/s = 1.2 kg(v) = (–0.4 kg m/s)/(1.2 kg) = v = –0.33 m/s
So, the cars are traveling at -0.33 m/s in the direction of the second car.
Hope this helps
<em>Tobey</em>
Answer:
Fc = 89.67N
Explanation:
Since the rope is unstretchable, the total length will always be 34m.
From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:
L1+L2=34m
Replacing this value in the previous equation:
Solving for H:

We can now, calculate the angle between L1 and the 2m segment:

If we make a sum of forces in the midpoint of the rope we get:
where T is the tension on the rope and F is the exerted force of 87N.
Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

Answer:
d. 37 °C
Explanation:
= mass of lump of metal = 250 g
= specific heat of lump of metal = 0.25 cal/g°C
= Initial temperature of lump of metal = 70 °C
= mass of water = 75 g
= specific heat of water = 1 cal/g°C
= Initial temperature of water = 20 °C
= mass of calorimeter = 500 g
= specific heat of calorimeter = 0.10 cal/g°C
= Initial temperature of calorimeter = 20 °C
= Final equilibrium temperature
Using conservation of heat
Heat lost by lump of metal = heat gained by water + heat gained by calorimeter
