Formation of an insoluble solid
Explanation:
One of the remarkable visible signs that indicates a precipitation reaction when two solutions are mixed is the formation of an insoluble solid. The insoluble solid formed is the precipitate.
- Precipitates usually forms in single replacement reactions and double replacement or double decomposition reactions.
- They form when two soluble compounds react. One of the product is an insoluble solid in the solution called the precipitate.
- The solubility table helps to predict whether precipitates forms in a reaction.
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Answer:
6 s
Explanation:
given,
Sports car accelerate from 0 to 30 mph in 1.5 s
time taken to accelerate 0 to 60 mph = ?
The power of the engine is independent of velocity and neglecting friction
power =
P = constant
the kinetic energy for 60 mph larger than this of 30 mph
= 
= 
= 
= 4
gain in kinetic energy = P x t
time = 4 x 1.5
= 6 s
As per the third law of Newton, the force exerted by the boat over the student is equal in magnitude to the force that the student exerted on the boat.
So, calculate the force on the student using the second law of Newton, Force = mass * acceleration.
Force on the student = 60 kg * 2.0 m/s^2 = 120 N.
=> horizontal force exerted by the student on the boat = 120 N
Answer: option d. 120 N. toward the back of the boat.
Of course it is toward the back because that is where the student jumped from..
We know that tangential acceleration is related with radius and angular acceleration according the following equation:
at = r * aa
where at is tangential acceleration (in m/s2), r is radius (in m) aa is angular acceleration (in rad/s2)
So the radius is r = d/2 = 1.2/2 = 0.6 m
Then at = 0.6 * 5 = 3 m/s2
Tangential acceleration of a point on the flywheel rim is 3 m/s2
Answer:
a) p = m1 v1 + m2 v2
, b) dp / dt = m1 a1 + m2 a2
, c) It is equivalent to force
dp / dt = 0
Explanation:
In this problem we have two blocks and the system is formed by the two bodies.
Part A. Initially they ask us to find the moment of the whole system
p = m1 v1 + m2 v2
Part B.
Find the derivative
dp / dt = m1 dv1dt + m2 dv2 / dt
dp / dt = m1 a1 + m2 a2
Part C.
Let's analyze the dimensions
m a = [kg] [m / s2] = [N]
It is equivalent to force
Part d
Acceleration is due to a net force applied
Part e
The acceleration of block 1 is due to the force exerted by block 2 during the moment change
Part f
Force of block 1 on block 2
True f12 = m1a1 f21 = m2a2
Part g
By the law of action and reaction are equal magnitude F12 = f21
Part H
dp / dt = 0
Isolated system F12 = F21 and the masses are constant. The total moment is only redistributed