The density is 81.4 g/m3. Before you start plugging numbers into the density formula (D=M/V), you should convert 104 kg to grams, which ends up being 104,000 grams. Then you can plug in the 104,000 grams and 1,278 m3 into the formula. When you divide the mass by the volume, you get a really long decimal, which you can round to 81.4 g/m3, or whatever place your teacher wants you to round to.
<span>To find the acceleration we are given two facts to begin. The impact at 16 km/h and the dent of 6.4 cm, or 0.064 meters. In solving the problem uniform acceleration is assumed, which would mean the avg speed during the impact was 8 km/hr by taking 16/2. We know distance = rate*time (d=r*t) . So t = d / r, so 0.64/8 = 0.008hr for t. Now we can solve for acceleration by taking a = 16 / 0.008 = 2000 km/hr.</span>
Answer:
I really hope this is right I think this is Diffuse I'm sorry if its worng
Answer:
Loss, 
Explanation:
Given that,
Mass of particle 1, 
Mass of particle 2, 
Speed of particle 1, 
Speed of particle 2, 
To find,
The magnitude of the loss in kinetic energy after the collision.
Solve,
Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.
Applying the conservation of momentum to find the speed of two particles after the collision.



V = 6.71 m/s
Initial kinetic energy before the collision,



Final kinetic energy after the collision,



Lost in kinetic energy,



Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.
Answer:
Option B. 8.1
Explanation:
From the question given above, the following data were obtained:
Angle θ = 71°
Hypothenus = 25
Adjacent = x
Thus, we can obtain the x component of the vector by using the cosine ratio as illustrated below:
Cos θ = Adjacent /Hypothenus
Cos 71 = x/25
Cross multiply
x = 25 × Cos 71
x = 25 × 0.3256
x = 8.1
Therefore, the x component of the vector is 8.1