The person does not sink into the snow because the force acts on a larger area so that the pressure is less
The edge of the sharp knife has a smaller area so the force acting on the knife produces a larger pressure
As the Earth rotates, it also moves, or revolves, around the Sun. ... As the Earth orbits the Sun, the Moon orbits the Earth. The Moon's orbit lasts 27 1/2 days, but because the Earth keeps moving, it takes the Moon two extra days, 29 1/2, to come back to the same place in our sky.
Answer:
attached below is the free body diagram of the missing illustration
Initial kinetic energy of the electron = 3 eV
Explanation:
The conclusion that can be drawn about the kinetic energy of the electron is

E
= initial kinetic energy of the electron
E
= -4 eV
E
= -1 eV
insert the values into the equation above
= -1 -(-4) eV
= -1 + 4 = 3 eV
Answer:
W = 30 J
Explanation:
given,
Work done = 10 J
Stretch of spring, x = 0.1 m
We know,
dW = F .dx
we know, F = k x


![W = k[\dfrac{x^2}{2}]_0^{0.1}](https://tex.z-dn.net/?f=W%20%3D%20k%5B%5Cdfrac%7Bx%5E2%7D%7B2%7D%5D_0%5E%7B0.1%7D)

k = 2000
now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.

![W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx](https://tex.z-dn.net/?f=W%20%3D%202000%20%5B%5Cdfrac%7Bx%5E2%7D%7B2%7D%5D_%7B0.1%7D%5E%7B0.2%7D%20dx)
W = 1000 x 0.03
W = 30 J
Hence, work done is equal to 30 J.
Answer:

Explanation:
<u>Motion in The Plane</u>
When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.
The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of
and
as the initial speed and angle, then we have




If we want to know the maximum height reached by the object, we find the value of t when
becomes zero, because the object stops going up and starts going down

Solving for t

Then we replace that value into y, to find the maximum height

Operating and simplifying

We have

The maximum height is

