Answer:
c. the tilt of the axis of rotation with respect to the Plane of the Ecliptic
Explanation:
The inclination of the ecliptic (or known only as obliqueness) refers to the angle of the axis of rotation with respect to a perpendicular to the plane of the eclipse. He is responsible for the seasons of the year that the planet Earth lends. It is not constant but changes through the movement of nutation. The terrestrial plane of Ecuador and the ecliptic intersect in a line that has an end at the point of Aries and at the diametrically opposite point of Libra.
When the Sun crosses the Aries, the spring equation occurs (between March 20 and 21, the beginning of spring in the northern hemisphere and the early autumn of the southern hemisphere), and from which the Sun is in the North Hemisphere; Pound until you reach the point of the autumn equinox (around September 22-23, beginning fall in the northern hemisphere and spring in the southern hemisphere).
Answer:
Option c is correct
Explanation:
There are two types of collisions-elastic collision and inelastic collision.
In elastic collision, both kinetic energy and total momentum are conserved. On the other hand, in inelastic collision, total momentum is conserved but kinetic energy is not conserved. Thus, option b and d are incorrect.
Total energy is always conserved in both types. Thus, option a is incorrect.
In a perfectly inelastic collision, objects stick together. This happens because maximum kinetic energy is dissipated and used in bonding of the two objects. Thus, correct option is c.
<h2>
Distance traveled in 1 second after drop is 4.9 m</h2><h2>
Distance traveled in 4 seconds after drop is 78.4 m</h2>
Explanation:
We have s = ut + 0.5at²
For a free falling object initial velocity u = 0 m/s and acceleration due to gravity, g = 9.8 m/s²
Substituting
s = 0 x t + 0.5 x 9.8 x t²
s = 4.9t²
We need to find distance traveled in 1 s and 4 s
Distance traveled in 1 second
s = 4.9 x 1² = 4.9 m
Distance traveled in 4 seconds
s = 4.9 x 4² = 78.4 m
Distance traveled in 1 second after drop = 4.9 m
Distance traveled in 4 seconds after drop = 78.4 m
Answer:
The new height the ball will reach = (1/4) of the initial height it reached.
Explanation:
The energy stored in any spring material is given as (1/2)kx²
This energy is converted to potential energy, mgH, of the ball at its maximum height.
If the initial height reached is H
And the initial compression of the spring = x
So, mgH = (1/2)kx²
H = kx²/2mg
The new compression, x₁ = x/2
New energy of loaded spring = (1/2)kx₁²
And the new potential energy = mgH₁
mgH₁ = (1/2)kx₁²
But x₁ = x/2
mgH₁ = (1/2)k(x/2)² = kx²/8
H₁ = kx²/8mg = H/4 (provided all the other parameters stay constant)
Acceleration = Change in Velocity / time
a = (v - u) / t
Where v = final velocity in m/s
u = initial velocity in m/s
t = time in seconds.
a = acceleration in m/s²
A proper record of the changes in velocity with the corresponding time would help find the acceleration.
