Answer:
428.59 N
Explanation:
Buoyant force, where V is volume, g is gravitational constant and \rho is density
where
is upward force
where
is the density of hippo
Using g as 9.81
Therefore, the upward force=428.59 N
Packages having a mass of 6 kgkg slide down a smooth chute and land horizontally with a speed of 3 m/sm/s on the surface of a co
1 answer:
Answer:
t = 1.02 s
Explanation:
The computation of the time required is shown below:
The package speed for belt is
= 3 - 1
= 2 m/s
Moreover, the decelerative force would be acted on the block i.e u.m.g
So, the decelerative produced
= 0.2 × 9.81
= 1.962 m/s^2
And, final velocity = 0
v = u - at
here
V = 0 = final velocity
u = 2 m/s
so,
0 = 2 - 1.962 × t
t = 1.02 s
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The position function x(t) of a particle moving along an x axis is
a) The point at which particle stop, it's velocity = 0 m/s
So dx/dt = 0
0 = 0- 12t = -12t
So when time t= 0, velocity = 0 m/s
So the particle is starting from rest.
At t = 0 the particle is (momentarily) stop
b) When t = 0
SO at x = 4m the particle is (momentarily) stop
c) We have
At origin x = 0
Substituting
t = 0.816 seconds or t = - 0.816 seconds
So when t = 0.816 seconds and t = - 0.816 seconds, particle pass through the origin.