The cat has two directions of motions:
The horizontal motion = Dx = 2.2 m
The vertical motion = Dy = -1.3 m (negative sign indicates that the cat is falling)
a = 9.8 m/sec^2
Vy = zero (since you are not moving up)
From the laws of motion:
<span>Dy = Vyt + 0.5ayt^2
</span>-1.3 = 0(t) + 0.5(-9.8)t^2
<span>t = 0.52s
</span>
Then, again using the laws of motion (but for the horizontal direction this time)
Dx = Vxt
<span>2.2 = Vx0.52 </span>
<span>Vx = 2.2/0.52 </span>
<span>= 4.23 m/s
</span>
<span>Therefore the cat's speed when it slid off the table is 4.23 m/s horizontally.</span>
Answer:
Speed of both blocks after collision is 2 m/s
Explanation:
It is given that,
Mass of both blocks, m₁ = m₂ = 1 kg
Velocity of first block, u₁ = 3 m/s
Velocity of other block, u₂ = 1 m/s
Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :
v = 2 m/s
Hence, their speed after collision is 2 m/s.
This is something u are going to have to do
Answer:
The correct answer is Option A (decrease).
Explanation:
- According to Heisenberg's presumption of unpredictability, it's impossible to ascertain a quantum state viewpoint as well as momentum throughout tandem.
- Also, unless we have accurate estimations throughout the situation, we will have a decreased consistency throughout the velocity as well as vice versa though too.
Other given choices are not connected to the given query. Thus the above is the right answer.
