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: 30 metres
Explanation:
Initial velocity of object = 120m/s
Time taken = 4.0s
Distance covered by object = ?
Recall that distance = (Change in velocity / Time taken)
Distance = (120m/s)/4.0s
= (120m/s) / 4.0s
= 30m
Thus, the object will be 30 metres high
Answer:
The height of the sound waves increases
Explanation:
Answer:
Work done, W = 2675.4 J
Given:
mass, m = 70.0 kg
height, H = 3.90 m
Solution:
According to the question, as the person jumps the stairs up, there is an increase in the potential energy of the person which is provided by the work done in climbing the stairs and is given by:
Work done, W = mgH
where
g = acceleration due to gravity = ![9.8 m/s^{2}[tex][tex]W = 70.0\times 9.8\times 3.90 = 2675.4 J](https://tex.z-dn.net/?f=9.8%20m%2Fs%5E%7B2%7D%5Btex%5D%3C%2Fp%3E%3Cp%3E%5Btex%5DW%20%3D%2070.0%5Ctimes%209.8%5Ctimes%203.90%20%3D%202675.4%20J)
