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>
<span>3.92 m/s^2
Assuming that the local gravitational acceleration is 9.8 m/s^2, then the maximum acceleration that the truck can have is the coefficient of static friction multiplied by the local gravitational acceleration, so
0.4 * 9.8 m/s^2 = 3.92 m/s^2
If you want the more complicated answer, the normal force that the crate exerts is it's mass times the local gravitational acceleration, so
20.0 kg * 9.8 m/s^2 = 196 kg*m/s^2 = 196 N
Multiply by the coefficient of static friction, giving
196 N * 0.4 = 78.4 N
So we need to apply 78.4 N of force to start the crate moving. Let's divide by the crate's mass
78.4 N / 20.0 kg
= 78.4 kg*m/s^2 / 20.0 kg
= 3.92 m/s^2
And you get the same result.</span>
I think it’s false. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and then mass of the object.
The answer is C (the same number of valence electrons)
work is distance * force so 15*100=1500
and to find time you know power = diastance * force / time
so 25=15*100/t
25=1500/t
25/1500=t
.016=time
