Answer:
30.36°
Explanation:
By using linear momentum; linear momentum can be expressed by the relation:
where ;
m= mass
= velocity of components in the x direction
= velocity of components in the y direction
If we consider the east as the positive x and north as positive y which is synonymous to what we usually have on a graph.
Then;
Initial momentum = 
= 
However, the masses stick together after collision and move with a common velocity: 
∴ Final momentum = 
= 
From the foregoing ;
initial momentum = final momentum
So;
Finally;
The required angle θ = 
θ = 
θ = 
θ = 30.36°
Answer:
Atomic size
Explanation:
In the periodic table , atomic size is indirectly proportional to the effective nuclear charge .the atomic size reduces from left to right across the table. This is because electrons are added to the same shell.
Answer:
the energy of the spring at the start is 400 J.
Explanation:
Given;
mass of the box, m = 8.0 kg
final speed of the box, v = 10 m/s
Apply the principle of conservation of energy to determine the energy of the spring at the start;
Final Kinetic energy of the box = initial elastic potential energy of the spring
K.E = Ux
¹/₂mv² = Ux
¹/₂ x 8 x 10² = Ux
400 J = Ux
Therefore, the energy of the spring at the start is 400 J.
A stone is thrown vertically upward with a speed of 17.0 m/s. How fast is it moving when it reaches a height of 11.0 m? How long is required to reach this height?
Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):
s = ut + ½ at^2 …. (1)
v^2 = u^2 + 2as …. (2)
v = u + at …. (3)
s = (u + v)t/2 …. (4)
where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.
In this case, we know u = 17.0m/s, a = -g = -9.81m/s^2, s = 11.0m and we want to know v and t, so from equation (2):
v^2 = u^2 + 2as
v^2 = 17.0^2 -2(9.81)(11.0)
v = √73.18 = 8.55m/s
now from equation (3):
v = u + at
8.55 = 17.0 – 9.81t
t = (8.55 – 17.0)/(-9.81) = 0.86s
The maximum current in an ac circuit is related to the rms current by:
where
is the rms current and
is the maximum current.
If we re-arrange the formula and we use
, we can find the value of the maximum current:
