John can run with the velocity of 5 m/s
Explanation:
- Kinetic energy is defined as the energy is being used to do an activity, basically energy associated with the motion of objects in the universe.
- The formula used to find the kinetic energy of an object is k = where as k represented as kinetic energy, m is the mass of the object and v is the velocity of the given object.
- Here, to find the answer we have to re-write the equation as
- Given, the mass of the object, here it is John = 80 kg, energy needs to be converted to kinetic energy, k = 1000 J.
- Hence, substitute all the values, then you would velocity as 5 m/s
Answer:
m = 8
Explanation:
A telescope is a device that allows us to see objects that were very far from us, it is built by the combination of two lenses, the one with the lowest focal length near the eye and that is the one or the one with the greatest focal length, the most eye-flounder . The magnification of the telescope is
m = - f₀ /
Where f₀ is the focal length of the lens and f_{e} is the false distance of the eyepiece.
It is this problem that gives us the diopter of each lens, these are related to the focal length in meters
D = 1 / f
Let's find the focal length
f₁ = 1 / D₁
f₁ = 1 / 1.16
f₁ = 0.862 m
f₂ = 1 / 9.37
f₂ = 0.1067 m
Therefore, the lens with f₂ is the eyepiece and the slow one with the
distance focal f₁ is the objective.
Let's calculate
m = - f₂ / f₁
m = - 0.862 / 0.1067
m = 8
The air drag is a force that depends on the speed of an object relative to the wind. Under certain conditions, it can be modeled as:
Where b is a constant.
As a falling object reaches a speed so that its weight is cancelled out by the air drag, the object will reach a maximum velocity.
In a speed vs time gaph, the speed would approach the maximum speed like an asymptote.
On the other hand, since the object falls from rest, the initial speed on the graph must be zero.
Taking these considerations into account, the correct graph for the movement of an object that falls from rest if air drag is not ignored, is option B.
Answer:
The speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.
Explanation:
Let u is the initial speed of the launch. Using first equation of motion as :
a=-g
The velocity of the shell at launch and 5.4 s after the launch is given by :
So, the speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.