Answer:
its probably still trying to load ur next rank or whatever it did it to me too
Explanation:
<u>ALL of the following work assumes NO AIR RESISTANCE:</u>
1). an object moving under the influence of only gravity, and not in orbit; its horizontal velocity is constant, and its vertical motion is accelerated downward at 9.8 m/s²
2). a parabola
3). Horizontal: velocity is constant, acceleration is zero. . . . Vertical: acceleration is 9.8 m/s² downward, velocity depends on whether it was launched, thrown up, thrown down, dropped, etc.
4). a). the one that was thrown horizontally; b). both hit the ground at the same time; c). both hit the ground with the same vertical velocity
5). a). zero; b). zero; c). gravity ... 9.8 m/s² down; d). 3.06 seconds; e). 4.38 m/s; f). 30 m/s g). no; gravity has no effect on horizontal motion
6). a). 1.8 seconds; b). 13.1 meters; c). 17.6 m/s down; d). 7.3 m/s; gravity has no effect on horizontal motion
7). 45 m/s
8). without air resistance, the ball is traveling horizontally at 13 km/hr, and it lands back in your hand
9). a). 4.49 m/s; b). 29.7 m/s
10). 7.24 meters
11). 700 meters
12). A). 103.7 meters ( ! she's in big trouble ! ); B). 17.5 meters
Answer:
2.06 m/s
Explanation:
From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.
Momentum before collision
Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5
Momentum after collision
The momentum after collision will be given by (9+27)*0.9=32.4
Relating the two then 9v+13.5=32.4
9v=18.5
V=2.055555555555555555555555555555555555555 m/s
Rounded off, v is approximately 2.06 m/s
Answer:
6200 J
Explanation:
Momentum is conserved.
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
The car is initially stationary. The truck and car stick together after the collision, so they have the same final velocity. Therefore:
m₁ u₁ = (m₁ + m₂) v
Solving for the truck's initial velocity:
(2700 kg) u = (2700 kg + 1000 kg) (3 m/s)
u = 4.11 m/s
The change in kinetic energy is therefore:
ΔKE = ½ (m₁ + m₂) v² − ½ m₁ u²
ΔKE = ½ (2700 kg + 1000 kg) (3 m/s)² − ½ (2700 kg) (4.11 m/s)²
ΔKE = -6200 J
6200 J of kinetic energy is "lost".
