D=vt use this equation to get the depth of the skull
Answer:
Option B
Explanation:
<h3>According to Newton's third law, for every reaction there will be equal and opposite reaction</h3>
Here in this case the force of the club hitting the golf ball will be in one direction and the force acting on club due to golf ball will be in opposite direction and magnitude of this force will be same as the magnitude of the force of the club hitting the golf ball
In this case the action will be the force of the club hitting the golf ball and reaction will be the force acting on club due to golf ball
∴ The club pushes against to golf ball with a force equal and opposite to the force of the golf ball on the club
Answer:
DOUBLE CHECK BECUASE IM ONLY 68.030303039999999% SURE!!!
(ANSWER IS HERE) ( D) It lacked practical examples in supporting theory
Know it's not B becuase there was no scientific community back then.
Know it's not C becuase it actully had lots of evidence.
But I'm not sure about A
Answer:
15.065ft
Explanation:
To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.
By definition the drag force is expressed as:

Where
is the density of the flow
V = Velocity
= Drag coefficient
A = Area
For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3
For second Newton's Law the Force is also defined as,

Equating both equations we have:



Integrating


Here,






Replacing:




Answer:
when the direction of the Current changes.
Explanation:
Electromagnet refers to an iron ore wrapped around with a coil of wire, in presence of electric current. As it acts like a magnet, when current is passed through it.
The north & south poles of magnetic fields produced by such magnet, change with direction of current passed through it.