Answer:
Total momentum = 16 Kgm/s
Explanation:
Let the momentum of the two balls be A and B respectively.
Momentum A = 16 kgm/s
Momentum B = 0 kgm/s (since the ball is at rest).
Total momentum = A + B
Total momentum = 16 + 0
Total momentum = 16 Kgm/s
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
Answer:
Approximately
.
Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.
Explanation:
Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant
near the surface of the earth.
For an object that is accelerating constantly,
,
where
is the initial velocity of the object,
is the final velocity of the object.
is its acceleration, and
is its displacement.
In this case,
is the same as the change in the ball's height:
. By assumption, this ball was dropped with no initial velocity. As a result,
. Since the ball is accelerating due to gravity,
.
.
In this case,
would be the velocity of the ball just before it hits the ground. Solve for
.
.
Answer:
355 m/s
Explanation:
Distance = 605 km
Initial speed =
= 284 m/s
Final velocity =
= 426 m/s
Average speed = ?
There is two method two find average speed. In first method, using 3rd equation of motion, we find acceleration.

Then using first equation of motion, we find time

Then using the formula of average velocity, we find average velocity

Second method is very simple


355 m/s
Answer:
E/4
Explanation:
The formula for electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
Where;
E is the electric field
σ is the surface charge density
ε₀ is the electric constant.
Formula to calculate σ is;
σ = Q/A
Where;
Q is the total charge of the sheet
A is the sheet's area.
We are told the elastic sheet is a square with a side length as d, thus ;
A = d²
So;
σ = Q/d²
Putting Q/d² for σ in the electric field equation to obtain;
E = Q/(2ε₀d²)
Now, we can see that E is inversely proportional to the square of d i.e.
E ∝ 1/d²
The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.
From the relationship of E with d, the magnitude of electric field at P will now have a quarter of its original magnitude which is;
E_new = E/4