It's total kinetic energy
Answer:
Newton's law of universal gravitation states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Explanation:
Answer:5
Explanation:
Given
speed of object ![v=1\ m/s](https://tex.z-dn.net/?f=v%3D1%5C%20m%2Fs)
radius of circle ![r=1\ m](https://tex.z-dn.net/?f=r%3D1%5C%20m)
Force towards the center ![F=1\ N](https://tex.z-dn.net/?f=F%3D1%5C%20N)
Work done is given by the dot product of Force and displacement
and we know know displacement of the object is along the circle which is perpendicular to the force acting therefore Work done will be zero
![W=F\cdot s\cos 90](https://tex.z-dn.net/?f=W%3DF%5Ccdot%20s%5Ccos%2090)
![W=0](https://tex.z-dn.net/?f=W%3D0)
Answer:
Explanation:
You pull a sled exerting a 50 N force on it , sled also exerts a force on you . These forces are action and reaction force , as per third law of Newton . These two forces are equal and opposite . But they do not act on the same object so they do not cancel each other . They act on different objects , one on the sledge and the other on you . Due to force on sledge , sledge moves in the direction of force or towards you . You will start moving in opposite direction if frictional force of ground is nil or less .
Answer:
Approximate height of the building is 23213 meters.
Explanation:
Let the height of the building be represented by h.
0.02 radians = 0.02 × ![\frac{180^{o} }{\pi }](https://tex.z-dn.net/?f=%5Cfrac%7B180%5E%7Bo%7D%20%7D%7B%5Cpi%20%7D)
= 0.02 x (180/
)
0.02 radians = 1.146°
10.5 km = 10500 m
Applying the trigonometric function, we have;
Tan θ = ![\frac{opposite}{adjacent}](https://tex.z-dn.net/?f=%5Cfrac%7Bopposite%7D%7Badjacent%7D)
So that,
Tan 1.146° = ![\frac{h}{10500}](https://tex.z-dn.net/?f=%5Cfrac%7Bh%7D%7B10500%7D)
⇒ h = Tan 1.146° x 10500
= 2.21074 x 10500
= 23212.77
h = 23213 m
The approximate height of the building is 23213 m.