<h2>
Answer: 1000 J</h2>
The Work
done by a Force
refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.
It should be noted that it is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter:
Now, when the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:
(1)
When they are not parallel, both directions form an angle, let's call it
. In that case the expression to calculate the Work is:
(2)
For example, in order to push the 200 N box across the floor, you have to apply a force along the distance
to overcome the resistance of the weight of the box (its 200 N).
In this case both <u>(the force and the distance in the path) are parallel</u>, so the work
performed is the product of the force exerted to push the box
by the distance traveled
. as shown in equation (1).
Hence:
>>>>This is the work
Red shift is one piece of observational evidence which supports the big
bang theory. The fact that galaxies are moving away from us means that
the universe was once much smaller, supporting the big bang theory which
states that all the matter in the universe once occupied a single point
with infinite density.
solution:
As Given plane is flying in east direction.
It throws back some supplies to designated target.
Time taken by the supply to reach the target =10 seconds
g = Acceleration due to gravity = - 9.8 m/s²[Taken negative as object is falling Downwards]
As we have to find distance from the ground to plane which is given by d.
d = 
=
meters
Distance from the ground where supplies has to be land to plane = Option B =490 meters
The heat form the Mylar outside her window may transfer the heat to the walls. Which then tanker the heat inside Lucia’s home.
Answer:
1.2cm
Explanation:
V=(2ev/m)^1/2
=(2*1.6*10^19 x2500/ 1.67*10^27)^1/2
=6.2x10^5m/s
Radius of resulting path= MV/qB
= 1.67*10^-27x6.92*10^6/1.6*10^-16 x0.6
=0.012m
=1.2cm