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:
m = 28.7[kg]
Explanation:
To solve this problem we must use the definition of kinetic energy, which can be calculated by means of the following equation.
where:
Ek = kinetic energy = 1800 [J]
m = mass [kg]
v = 11.2 [m/s]
![1800=\frac{1}{2}*m*(11.2)^{2}\\m = 28.7[kg]](https://tex.z-dn.net/?f=1800%3D%5Cfrac%7B1%7D%7B2%7D%2Am%2A%2811.2%29%5E%7B2%7D%5C%5Cm%20%3D%2028.7%5Bkg%5D)
Answer:222 meters
Explanation:
Speed=80kph=(80x5/18)=(80x5)/18=22.2m/s
Time=10 seconds
distance=speed x time
distance=22.2 x 10
distance=222
distance=222 meters
Yes I would expect them too
Let the mass of 2500 kg car be 
and it's velocity be 
and the mass of 1500 kg car be 
and it's velocity be 
.
After the bumping the mass be M and it's velocity be V.
By law of conservation of momentum we have
2500 * 5 + 1500 * 1=4000 * V
V = 14000/4000 = 7/2 = 3.5 m/s
So the velocity of the two-car train = 3.5 m/s
