As we know that power is defined as rate of work done
so we will have

so in order to increase the power as per above formula we know that either we need to increase the work or we need to decrease the time to complete that work
So here the correct answer will be
increase the work being done or decrease the time in which the work is completed.
The speed of a proton after it accelerates from rest through a potential difference of 350 V is
.
Initial velocity of the proton 
Given potential difference 
let's assume that the speed of the proton is
,
Since the proton is accelerating through a potential difference, proton's potential energy will change with time. The potential energy of a particle of charge
when accelerated with a potential difference
is,

Due to Work-Energy Theorem and Conservation of Energy - <em>If there is no non-conservative force acting on a particle then loss in Potential energy P.E must be equal to gain in Kinetic Energy K.E</em> i.e

If the initial and final velocity of the proton is
and
respectively then,
change in Kinetic Energy 
change in Potential Energy 
from conservation of energy,

so, 

To read more about the conservation of energy, please go to brainly.com/question/14668053
Answer:
This depends on the writers
if they want they can make spiderman deny the laws of nature
Answer:
demand (an amount) as a price from someone for a service rendered or goods supplied.
<h2>5.3 km</h2>
Explanation:
This question involves continuous displacement in various directions. When it becomes difficult to imagine, vector analysis becomes handy.
Let us denote each of the individual displacements by a vector. Consider the unit vectors
as the unit vectors in the direction of East and North respectively.
By simple calculations, we can derive the unit vectors
in the directions North,
South of West and
North of West respectively.
So Total displacement vector = Sum of individual displacement vectors.
Displacement vector = 
Magnitude of Displacement = 
∴ Total displacement = 