Answer:
the cat is 0.4238 m in front of the dog as it leaps through the window
Explanation:
Given that;
acceleration a = 0.85 m/s²
speed v = 1.40 m/s
the cat is at rest, so initial velocity u = 0
we know that, since the cat is sleeping on the floor in the middle of a 2.8-m-wide room, it needs to cover (2.8 m / 2 ) distance to get to the window;
using the second equation equation of motion;
s = ut + 1/2 at²
we substitute
2.8/2 = 0×t + 1/2 × 0.85 × t²
1.4 = 0.425t²
t = √( 1.4 / 0.425 )
t = 1.81497 sec
now, at acceleration 0.10 m/s²
the dog has to cover the distance;
s = ut + 1/2 at²
s = ( 1.4 × 1.81497) - 1/2 × 0.10 × 1.81497²
s = 2.540958 - 0.1647
s = 2.3762 m
The cant in front of the dog as it leaps through the window;
distance = 2.8 m - 2.3762 m
distance = 0.4238 m
Therefore, the cat is 0.4238 m in front of the dog as it leaps through the window
Answer:
560 m
Explanation:
The speed of sound in air is approximately:
v ≈ v₀ + 0.6T
where v₀ is the speed of sound at 0°C (273 K) in m/s, and T is the temperature in Celsius.
The speed of sound at 20°C at that altitude is:
v ≈ 327 + 0.6(20)
v ≈ 339 m/s
The sound travels from the hikers to the mountain and back again, so it travels twice the distance.
339 m/s = 2d / 3.3 s
2d = 1118.7 m
d = 559.35 m
Rounding, the mountain is approximately 560 m away.
Water cycle, evaporation, condensation, and freezing
