Complete Question
A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.
Answer:
The value 
Explanation:
From the question we are told that
The charge on the first sphere is 
The charge on the second sphere is 
The mass of the second charge is 
The distance apart is 
The speed of the second sphere is 
Generally the total energy possessed by when
and
are separated by
is mathematically represented

Here KE is the kinetic energy which is mathematically represented as

substituting value


And U is the potential energy which is mathematically represented as

substituting values


So


Generally the total energy possessed by when
and
are separated by
is mathematically represented

Here
is the kinetic energy which is mathematically represented as

substituting value


And
is the potential energy which is mathematically represented as

substituting values


From the law of energy conservation

So


Answer:
Technique of comparing abundance ratio between radioactive isotopes to a reference isotope to determine the age of a material called radioactive dating. It determines the age by having a more abundance of isotopes in the cellular being.
Answer:
51.96 m/s^-1
Explanation:
a) see the attachment
b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be
v_ox=v_o*cosФ
=60*cos (30)
= 51.96 m/s^-1
Answer:
The gravitational force changing velocity is

Explanation:
The expression for the gravitational force is

Differentiate the above equation

The velocity is the distance in at time so



The efficiency of an ideal Carnot heat engine can be written as:

where

is the temperature of the cold region

is the temperature of the hot region
For the engine in our problem, we have

and

, so the efficiency is